Complex Analysis.pdf

Chapter 1

Preliminaries (1) arg(z1 .z2 . . . . .zn ) = arg(z1 ) + ... + arg(zn ) (2) arg(z n ) = narg(z) (3) arg( zz12 ) = arg(z1 ) − arg(z2 ) CONCEPTS OF ANALYTIC FUNCTIONS 1. Limits and continuity: In complex numbers, we know that there are 4 kinds of functions (1) Real valued functions of a real variable. (2) Real valued function of a complex variable. (3) Complex valued function of a real variable. (4) Complex valued function of a complex variable. 2. Limit of a function: The function f (x) has a limit A as x → a lim f (x) = A

x→a

(1.1)

if and only if for every  > 0 ∃ δ > 0 with the property that |f (x) − A| <  for all values of x such that |x − a| < δ. and x 6= a Results: 1

• lim f (x) = A and lim g(x) = B x→a

x→a

• lim [f (x) + g(x)] = lim f (x) + lim g(x) = A + B x→a

x→a

x→a

• lim [f (x) − g(x)] = lim f (x) − lim g(x) = A − B x→a

x→a

x→a

• lim [f (x)g(x)] = lim f (x). lim g(x) = AB x→a

x→a

x→a

• lim [f (x)/g(x)] = A/B where g(x) 6= 0∀x x→a

Now (1.1) is equivalent to lim f (x) = A

(1.2)

x→a

and lim Re[f (x)] = Re(A)

x→a

lim Imf (x) = ImA

x→a

(1.3)

Hence from (1.1) and (1.3), we have lim f (x) = A iff lim Re(f (x)) = Re(A) and lim Im(f (x)) = Im(A). Consider

x→a

x→a

x→a

the transformation w = f (z) = 1/z. At z = 0 is mapped to w = ∞ in w-plane similarly by z = ∞ is mapped into w = 0 the point at ∞ in z-plane. Continuous function at a point: The function f (x) is continuous at a point 0 a0 if and only if (1) lim f (x) exists. x→a

(2) f (a) exist. (3) lim f (x) = f (a) x→a

for  > 0 ∃ δ > 0 such that |f (x) − f (a)| <  whenever |x − a| < δ. Continuous function in a region:

2

A function is continuous in a region if it is continuous at all points in the region. Result: If f (x) and g(x) are continuous at x = a, we have: (1) f (x) + g(x) is continuous at x = a (2) f (x) − g(x) is continuous at x = a (3) f (x).g(x) is continuous at x = a (4) f (x)/g(x) is continuous at x = a when g(x) 6= 0 ∀x Examples: In every finite region, the following are continuous. (1) All polynomials in a finite region is continuous (2) exp z is continuous everywhere (3) cos z, sin z are continuous (4) w = f (z) is continuous at z = a, where z = g(ζ) is continuous at ζ = b then w = f [g(ζ)] is continuous at b continuous function of a continuous function is continuous If f (z) is continuous at z = a, then Re[f (z)] is continuous and im[f (z)] is continuous at z = a. Hence f (z) is continuous at z = a m Re[f (z)] and Im[f (z)] are continuous at z = a (5) f (z) = Re[f (z)] + iIm[f (z)] is continuous at z = a, then p |f (z)| = [(Ref (z)]2 + [Imf (z)]2 is also continuous at z = a. Derivative at a point :

3

The derivative of a function f (z) at a point z = (z0 ) is defined as, lim

z→z0

f (z) − f (z0 ) = f 0 (z0 ) z − (z0 )

Otherwise f (z + 4(z) − f (z) = f 0 (z) 4z→0 4(z) lim

provided that 0 lim0 exist independent of manner on which 4(z) → 0 Results: If f (z) and g(z) are differentiable at z = z0 then (1) f (z) + g(z) also differentiable at z = z0 (2) f − g differentiable at z = z0 (3) f.g differentiable at z = z0 (4) f /g is differentiable at z = z0 when g(z) 6= 0∀z (5) d/dz[f (z) + g(z)] = d/dz(f (z)) + d/dz(g(z)) (6) d/dz[f (z) − g(z)] = d/dz(f (z)) − d/dz(g(z)) (7) d/dz[cf (z)] = c(d/dz)f (z), c-constant (8) d/d[f (z)/g(z)] = [(g(z)(d/dz)f (z) − f (z)(d/dz)g(z))/[g(z)]2 ], g(z) 6= 0 (9) w(z) = f (g(z))0 (d/dz)(w) = d/dz(f (g(z)) = f 0 (g(z))g 0 (z) There is a fundamental difference between case of real and complex variable. Here we focus on complex valued function of a real variable and complex valued function of complex variable. Let f (z) be a complex variables real valued function whose derivative exist at z = a then f 0 (a) is one side real for it is the limit of quotients [(f (a+h)−f (a))/h] as h → 0 through real on the other side, it is also limit of (f (a+ih)−f (a)/ih) therefore 4

f 0 (a) = 0, thus real valued function of complex variable either has derivative 0 or derivative does not exist. Complex valued function of real variable : z(t) = x(t) + iy(t) ⇒ z 0 (t) = x0 (t) + iy 0 (t) Class of Analytic (Holomorphic) functions : The class of analytic function is formed by complex valued of complex variable which possess a derivative wherever the function is defined. The term holomorphic is used with similar meaning. The definition of derivative can be written in the form: f (z + h) − f (z) h→0 h

f 0 (z) = lim As a consequence (1) f (z) is continuous.

(2) If we write f (z) = u(z) + iv(z), it follows that u(z) and v(z) are continuous. The limit of difference quotient must be the same regardless of the wave in which h approaches 0. If we choose real values for h, then imaginary part y is constant, the derivative becomes a partial derivative with respect to x. Therefore, we have f 0 (z) =

∂u ∂v ∂f = +i ∂x ∂x ∂x

(1.4)

Similarly, if we purely imaginary value h = ik, we have f (z + ik) − f (z) h→0 ik f (z + ik) − f (z) = lim k→0 ik f (z + ik) − f (z) = lim −i k→0  ik  ∂f ∂u ∂v = −i = −i +i ∂y ∂y ∂y

f 0 (z) = lim

f 0 (z) =

∂v ∂u −i ∂y ∂y 5

(1.5)

from ( 1.4) and (1.5) ∂u ∂v ∂v ∂u = and =− ∂x ∂y ∂x ∂y

(1.6)

These are the Cauchy-riemann differential equation which must be satisfied by Re&Im part of analytic function. Remark 1. The existence of four partial derivatives in (1.6) is implied by existence of f 0 (z). Using (1.6) we can write, four formally different expression for f (z), the simplest is ∂v ∂u +i ∂x ∂x  2  2 ∂v ∂u + = ∂x ∂x ∂u ∂v ∂u ∂v = . + . ∂x ∂x ∂x ∂x

f 0 (z) = |f 0 (z)|2

Applying C-R equation |f 0 (z)|2 =

∂u ∂v ∂v ∂u . − . ∂x ∂y ∂x ∂y

∂u ∂v ∂x ∂y |f (z)| = ∂u ∂v is the Jacobian of x and y. ∂x ∂y A function u which satisfies Laplace equation is said to be Harmonic. The real and 0

2

imaginary part of analytic function are thus harmonic. If two harmonic functions u, v satisfy C-R equations, then v is said to be conjugate harmonic function of u ∂2 ∂2 + ∂x2 ∂y 2     ∂ 2u ∂ 2u ∂ ∂u ∂ ∂u ∆u = + = + ∂x2 ∂y 2 ∂x ∂x ∂y ∂y     ∂ ∂v ∂ ∂v ∂v ∂v = + − = − ∂x ∂y ∂y ∂x ∂x∂y ∂y∂x ∆u = 0 (since they are continuous). ∆ =

Similarly ∆v = 0. 6

Definition: 1 (Analytic functions). (1) If u(x, y) and v(x, y) has continuous first order partial derivatives and satisfy C-R equation, then f (z) = u(z) + iv(z) is analytic with continuous derivatives f 0 (z) and conversely. Example: If u = x2 − y 2 , then it is easy to see that u is harmonic since ∂u ∂u = 2x & = −2y ∂x ∂y ∂ 2u ∂ 2u = 2 & = −2 ∂x2 ∂y 2 ∂ 2u ∂ 2u ∆u = + =2−2=0 ∂x2 ∂y 2 ∴ u is harmonic. Assume v is the conjugate of u(harmonic). By C-R equations ∂v ∂u = = 2x ∂x ∂y

(1.7)

∂v ∂u =− = 2y ∂x ∂y

(1.8)

Integrating (1.8 ) with respect to x, Z v =

2ydx

v = 2xy + φ(y) ∂v ⇒ = 2x + φ0 (y) = 2x ∂y 0 φ (y) = 0 φ(y) = c ⇒ v = 2xy + c, u = x2 − y 2 f (z) = u + iv = x2 − y 2 + 2xyi + ic = z 2 + ic.

7

Remark 2. Consider a complex function f (x, y) of two real variables. Introducing complex variables z = x + iy and its conjugate z¯ = x − iy. Then z + z¯ 2 z − z¯ z − z¯ = 2iy ⇒ y = 2i z + z¯ = 2x ⇒ x =

With this change of variable we can consider f (x, y) as function of z and z¯. ∂f ∂z

∂f ∂x ∂f ∂y . + . ∂x ∂z ∂y ∂z   1 ∂f 1 ∂f ∂f 1 ∂f ∂f = + ⇒ = −i 2 ∂x 2i ∂y ∂z 2 ∂x ∂y ∂f ∂f ∂x ∂f ∂y 1 ∂f 1 ∂f = . + . = − ∂ z¯ ∂x ∂ z¯ ∂y ∂ z¯ 2 ∂x 2i ∂y 1 ∂f ∂f ∂f = ( +i ) ∂ z¯ 2 ∂x ∂y =

By using C-R equations, we get for f = u + iv and ⇒

∂f ∂x

= −i ∂f ∂y

∂f ∂f = = 0 ∂x ∂y ∂f ∴ = 0 ∂ z¯

We now remark that f (¯ z ) = f¯(z) = u(z) − iv(z) should have zero partial derivative with respect of z. This is because writing f = u + iv and f = u − iv. It is clear that ∂ f¯ ∂x

=

∂f ∂x

and

∂ f¯ ∂y

=

∂f . ∂y

Similarly

∂ f¯ ∂ f¯ ∂x ∂ f¯ ∂y = . + . ∂z ∂x ∂z ∂y ∂z 1 ∂ f¯ 1 ∂ f¯ . + . = 2 ∂x 2i ∂y ∂ f¯ ∂ f¯ ∂x ∂ f¯ ∂y 1 ∂ f¯ 1 ∂ f¯ = + + . = . − . ∂ z¯ ∂x ∂ z¯ ∂y ∂ z¯ 2 ∂x 2i ∂y We know ∂ f¯ ∂f = ( ) = ¯0 = 0 ∂x ∂ z¯ Hence in the variables z and z¯, f (z) can be considered purely as a function of z¯ and we denoted it by 8

f¯(¯ z ) = f (¯ z ), now we can write f (z) = u(x, y) + i[v(x, y)] ⇒ f (z) = u(x, y) − iv(x, y) f (z) + f (z) f (z) − f (z) , v(x, y) = 2 2i ∼ 1 [f (x + iy) + f (x − iy)] u(x, y) = 2 u(x, y) =

Substituting z z ,y = 2 2i ∼ 1 u(x, y) = [f (z) + f (0)] 2 x =

Since f (x) need to be determined up to a purely imaginary constant, we can assume that f (0) is real which implies ∼

f (0) = f (0) = f (0) = u(0, 0) ∴ f (z) has an explicit formula z z f (z) = 2u( , ) − u(0, 0)i(c) 2 2i Since a purely imaginary constant can be added without requirement u = Re(f ) Problem: 1. If g(w) and f (z) are analytic, show that g[f (z)] is also analytic Let g(w) = r(w)+is(w), f (z) = u(z)+iv(z) be analytic and Here z = x+iy,w = u+iv. To show g(f (z)) = r(w)+is(w) is analytic since r and s satisfy C-R equation in w ∂r ∂s ∂r ∂s = , =− ∂u ∂v ∂v ∂u

(1.9)

∂u ∂v ∂u ∂v = , =− ∂x ∂y ∂y ∂x

(1.10)

and since f is analytic

9

To prove that ∂r ∂s ∂r ∂s = , =− ∂x ∂y ∂y ∂x consider ∂r ∂u ∂r ∂v ∂r = . + . ∂x ∂u ∂x ∂v ∂x ∂s ∂s ∂u ∂s ∂v = . + . ∂y ∂u ∂y ∂v ∂y ∂r ∂s ∂r ∂u ∂r ∂v ∂s ∂u ∂s ∂v − = . + . − . − . ∂x ∂y ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y By using (1.9 ) and (1.10 ) ∂r ∂s ∂r ∂u ∂r ∂v ∂r ∂v ∂r ∂u − = . + + (− ) + ( ) ∂x ∂y ∂u ∂x ∂v ∂x ∂v ∂x ∂u ∂x ∂r ∂s − = 0 ∂x ∂y and ∂r ∂u ∂r ∂v ∂r = . + . ∂y ∂u ∂y ∂v ∂y ∂s ∂u ∂s ∂v ∂s = . + . ∂x ∂u ∂x ∂v ∂x ∂s ∂r ∂u ∂r ∂v ∂s ∂u ∂s ∂v ∂r + = . + . + . + . ∂y ∂x ∂u ∂y ∂v ∂y ∂u ∂x ∂v ∂x ∂r ∂u ∂u ∂r ∂v ∂r ∂v ∂r ∂u = . + + . − . + .(− ) ∂u ∂y ∂y ∂v ∂y ∂v ∂y ∂u ∂y = 0 Therefore, g(f (z)) is analytic. 2. Verify C-R equation for z 2 and z 3

f (z) = z 2 = (x + iy)2 = x2 − y 2 + i2xy u = x2 − y 2 , v = 2xy ∂u ∂v = 2x, = 2x; ∂x ∂y ∂u ∂v = ∂x ∂y

∂v ∂u = 2y, = −2y ∂x ∂y ∂u ∂v =− ∂y ∂x 10

Therefore, z 2 is analytic. f (z) = z 3 = (x + iy)3 = x3 − iy 3 + i3xy 2 − 3xy 2 u = x3 − 3xy 2 ,

v = 3x2 y − y 3

∂u ∂u = 3x2 − 3y 2 , = −6xy, ∂x ∂y ∂u ∂v ∂v ∂u ∴ = ; =− ∂x ∂y ∂x ∂y

∂v = 6xy, ∂x

∂v = 3x2 − 3y 2 ∂y

Therefore, z 3 is analytic. 3. Show that an analytic function cannot have a constant absolute value without reducing to a constant. Let f (z) = u + iv be an analytic function. Then it satisfy the C-R equations ux = vy ; vx = −uy Given |f (x)| = k(constant) ⇒

√

u2 + v 2 = k

u2 + v 2 = k 2 = c CASE I: Suppose c = 0. ⇒ u2 + v 2 = 0 ⇒ u2 = 0, v 2 = 0 ⇒ u = 0, v = 0. Therefore, f is a constant function. CASE 2: Suppose c 6= 0. u2 + v 2 = c 6= 0 11

(1.11)

Differentiating partially with respect to x, we get 2uux + 2vvx = 0 uux + vvx = 0

(1.12)

Differentiating partially with respect to y, we get 2uuy + 2vvy = 0 uuy + vvy = 0

(1.13)

Since u and v satisfy C-R equations (1.12) ⇒ uux − vuy = 0

(1.14)

(1.13) ⇒ uuy − vux = 0

(1.15)

(1.14)2 + (1.15)2 ⇒ u2 (ux )2 + v 2 (uy )2 − 2uvux uy + u2 (uy )2 + v 2 (ux )2 + 2uvux uy = 0 u2x [u2 + v 2 ] + u2y [u2 + v 2 ] = 0 (u2 + v 2 )(u2x + u2y ) = 0 From (1.11) (u2x + u2y ) = 0

⇒ u2x = 0 and u2y = 0

∴ ux = 0 and vy = 0 and u = c ⇒ v = c1 . ⇒ f is a constant function. z ) are simultaneously analytic 4. Prove that f (z) and f (¯ Suppose f (z) = u(x, y) + iv(x, y) is an analytic function. The first order partial derivatives u and v are continuous and satisfy C-R equation ux = vy

and uy = −vx

Then, f (¯ z ) = u(x, −y) + iv(x, −y) f (¯ z ) = u(x, −y) − iv(x, −y) 12

To prove: f (¯ z ) is analytic. f (¯ z ) = u1 (x, y) − iv1 (x, y) where u1 = u(x, −y),

v1 = −v(x, −y) ∂u ∂u1 = ; ∂x ∂x ∂u ∂u1 = − ; ⇒ ∂y ∂y

∂v1 ∂v = ∂x ∂x ∂v1 ∂v = ∂y ∂y

⇒

Therefore, ∂u1 ∂u ∂v ∂v1 = = = ∂x ∂x ∂y ∂y ∂u1 ∂u ∂v ∂v1 and = − = =− ∂y ∂y ∂x ∂x ∂u1 ∂v1 ∂u1 ∂v1 ⇒ = and =− ∂x ∂y ∂y ∂x ∴ u1 and v1 satisfy C-R equation. z ) is analytic. ∴ f (¯ 5. Prove that the function u(z) and u(¯ z ) are simultaneously harmonic. Suppose u(z) is harmonic and hence it satisfies the Laplace equation. ∂ 2u ∂ 2u + = 0 ∂x2 ∂y 2 u(¯ z ) = u(x, −y) = u1 (x, y). To prove ∂ 2 u1 ∂ 2 u1 + =0 ∂x2 ∂y 2

i.e., u1 is harmonic.

∂u1 ∂u ∂ 2 u1 ∂ 2u = ⇒ = ∂x ∂x ∂x2 ∂x2 2 ∂u1 ∂u ∂ u1 ∂ 2u and = − = = ∂y ∂y ∂y 2 ∂y 2 ∂ 2 u1 ∂ 2 u1 ∂ 2u ∂ 2u ∴ + = + =0 ∂x2 ∂y 2 ∂x2 ∂y 2 13

since u(x, −y)

∴ u1 is harmonic. (i.e.,) u(¯ z ) is harmonic.

6. Show that harmonic function satisfies the formal differential equation

∂2u ∂z∂ z¯

=0

Let u(z) = u(x + iy). we have   ∂ ∂u ∂ 2u = ∂z∂ z¯ ∂z ∂ z¯   1 ∂ ∂u ∂u = +i 2 ∂z ∂x ∂y   2 ∂ 2 u ∂x ∂ 2 u ∂y ∂ 2 u ∂y 1 ∂ u ∂x +i . + + i 2. = 2 ∂x2 ∂z ∂x∂y ∂z ∂x∂y ∂z ∂y ∂z   2  2 2 2 1 1 ∂ u ∂ u ∂ u ∂ u = +i + +i 2 2 2 2 ∂x ∂x∂y ∂x∂y ∂y  2  2 2 1 ∂ u ∂ u ∂ u ∂ 2u = +i −i − i(i) 2 4 ∂x2 ∂x∂y ∂x∂y ∂y  2  2 1 ∂ u ∂ u = + =0 4 ∂x2 ∂y 2 7. Show that u = x3 − 3xy 2 + 3x2 − 3y 2 + 1 is a harmonic function and find the corresponding analytic function. Then, ∂u = 3x2 − 3y 2 + 6x ∂x ∂ 2u = 6x + 6 ∂x2

∴

and and

∂u = −6xy − 6y ∂y ∂ 2u = −6x − 6 ∂y 2

∂ 2u ∂ 2u + = 6x + 6 − 6x − 6 = 0 ∂x2 ∂y 2

By C-R equation ,

and

∂u ∂v = = 3x2 − 3y 2 + 6x ∂x ∂y ∂u ∂v ∂v = − ⇒ = 6xy + 6y ∂y ∂x ∂x 14

(1.16)

Integrating (1.16 ) with respect to y, Z v = 3x2 − 3y 2 + 6xdy = 3x2 y − y 3 + 6xy + φ(x) ∂v = 6xy + 6y + φ0 (x) = 6xy + 6y ∂x ⇒ φ0 (x) = 0 ∴ φ(x) = c ∴ v = 3x2 y − y 3 + 6xy + c f (z) = u + iv f (z) = (x3 − 3xy 2 + 3x2 − 3y 2 + 1) + i(3x2 y − y 3 + 6xy + c) 8. Find most general harmonic polynomial of the form ax3 + bx2 y + cxy 2 + dy 3 . Determine conjugate harmonic function and corresponding analytic function by integration. Let u = ax3 + bx2 y + cxy 2 + dy 3 ∂u = 3ax2 + 2bxy + cy 2 ∂x ∂ 2u = 6ax + 2by ∂x2

and and

∂u = bx2 + 2cxy + 3y 2 d ∂y ∂ 2u = 2cx + 6yd ∂y 2

Since u is harmonic ∂ 2u ∂ 2u + = 0 ∂x2 ∂y 2 6ax + 2by + 2cx + 6yd = 0

x(6a + 2c) + y(2b + 6d) = 0 ∂u ∂v = = 3ax2 + 2bxy + cy 2 ∂x ∂y ∂u ∂v = = −bx2 − 2cxy − 3y 2 − ∂y ∂x 15

(1.17) (1.18) (1.19)

Integrating (1.18) partially with respect to y, v = 3ax2 y + bxy 2 +

cy 3 + φx 3

∂v = 6axy + by 2 + φ0 (x) = −bx2 − 2cxy − 3y 2 d ∂x φ0 (x) = −bx2 − 2cxy − 3y 2 d − 6axy − by 2

(1.20)

From(1.17 ) 6a = −2c 3a = −c

and b = −3d

2b = −6d c = −3a

( 1.20) becomes

φ0 (x) = −bx2 + 6axy − 3y 2 (−b/a) − 6axy − by 2 = −bx2 bx3 φ(x) = − 3 v = 3ax2 y + bxy 2 +

cy 3 bx3 − 3 3

f = u + iv   cy 3 bx3 2 2 = (ax + bx y + cxy + dy ) + i 3ax y + bxy + − 3 3 3

2

2

3

Definition: 2 (Polynomials). A polynomial in z is an expression of the form P (z) = a0 + a1 z + a2 z 2 + ..... + an z n where a0i s-complex numbers, zi -complex variable. Note: (1) We note that the identify function 0 z 0 is a non-constant analytic function in whole complex plane with derivative 1 and the function z n (n is positive) is analytic in the whole complex plane with derivative nz (n−1) . Since the sums and product of analytic functions are analytic, then, P (z) is analytic in whole complex plane with derivative P 0 (z) = a1 + 2a2 z + .... + nan z n−1 16

• In the above notation, we shall always assume an 6= 0and call P (z) as a polynomial of degree n. • We call α as a zero of P (z) if p(α) = 0. In the case by division algorithm we can find that P (z) = (z − α)P1 (z) where P1 (z): polynomial of degree n-1. Repeating the process, we can factorise P (z) as P (z) = (z − α1 )(z − α2 )..(z − αn ) where αi are not necessary distinct. Suppose h of αj0 s coincide, the value (αj ) is zero of order h for P (z). The order of a zero α of p(z) can also be determined by : P (α) = P 0 (α) = . . . = P h−1 (α) = 0 and P h (α) 6= 0. In other words, the order of zero α equals the order of the first non-vanishing derivative at α. A zero of order 1 is called a simple zero. Theorem: 1 (Lucas’s Theroam:). If all zeros of polynomial p(z) lie in a half-plane then, all zero of derivative p0 (z) lie in same half-plane. Proof. Let the factorisation of a polynomial be given by p(z) = an (z − α1 )(z − α2 ) . . . (z − αn ) where αi are not necessarily distinct the hypotheses implies that
Im αib−a < 0 for i ≤ i ≤ n. Suppose z is not in the half plane. in which αj0 s lie, Then,  Im

 z−a ≥ 0 b P (z) = an (z − α1 )(z − α2 ) . . . (z − αn ) P (z) = an (z − α2 ) . . . (z − αn ) + (z − α1 )(z − α2 ) . . . (z − αn ) + (z − α1 ) . . . (z − α(n−1) ) n

X 1 P 0 (z) 1 1 1 = + + ... + = P (z) z − α1 z − α2 z − αn (z − αi ) i=1

17

(1.21)

 Im

z − αi b



 = Im

     z − a + aαi z−a αi − a = Im − Im b b b   b ⇒ Im <0 for 1≤i≤n z − αi

from (1.21 ) n

bP 0 (z) X b = P (z) (z − αi ) i=1  0  (z) <0 ∴ Im b PP (z) and hence P 0 (z) 6= 0 If z is not in same in half plane then P 0 (z) 6= 0.
Then all zero of P 0 (z) lie in the same half plane given by Im z−a <0 b Definition: 3 (Rational Functions). A rational function is a function of the form R(z) =

P (z) Q(z)

where P (z) and Q(z) are two polynomial having no common factors,

that is, R(z) has no common zeros. Note: • By definition, when R(z) = ∞ the zeros of Q(z) which are called the ’poles of R(z). In general any rational function has finite zeros are the zeros of P (z) and finite poles which are zeros of Q(z). • If ’a’is a finite zero of R, then order of zero for R(z) at’a’is same as the order of zero for R(z) at’a’. • similarly, if ’b’is a finite pole for R(z) then order of pole for R(z) at b ,is same as order of zero of Q(z) at b. • To determine the nature of R(z) at ∞ , we shall define a new function, R1 (z) = R( z1 ) which is also rational P ( z1 ) 1 ⇒ R1 (z) = 1 = R( ) z Q( z )

18

SetR1 (0) = R(∞). Suppose R1 (0) = 0 → R(∞) = 0 =

P (∞) . Q(∞)

Therefore ∞ is a zero

for R(z) and the order of zero at ∞ for R(z) order of zero for R1 (z) at 0. Similarly If R1 (0) = R(∞) = ∞. ∴ ∞ is a pole for R(z) and its order for R(z) is same as the order of the pole at 0 for R1 (z). If R1 (0) 6= 0(or) ∞, R(z) has neither a pole (nor)zero at ∞. The derivative for R(z) is R0 (z) =

Q(z)P 0 (z)−P (z)Q0 (z) . (Q(z))2

when Q(z) 6= 0.

Result: • A rational function R(z) has sane number of zero and poles including those at ∞ and is equal to P , the maximum of m,n where m and n are degree of Q(z) and P (z) respectively. • This number P: order of rational function. And for any complex number ’a’the numerical solutions of equation R(z) = a = P Result: LetR(z) =

P (Z) , Q(z)

where P (z) = a0 + a1 z + a2 z 2 + .. + an z n and A(z) = b0 + b1 z +

b2 z 2 + .. + bm z m , an 6= 0, bm 6= 0 By fundamental theorem of Algebra, R(Z) has at most n finite zeros and m finite poles. We have to calculate the number of zeros or poles at ∞ for R(z).

  a0 + a1 ( z1 ) + a2 ( z12 ) + .. + an ( z1n ) 1 = R1 (z) = R z b0 + b1 ( z1 ) + b2 ( z12 ) + ...bm ( z1m ) z m a0 z n + a1 z n−1 + .. + an = [ ] z n b0 z m + b1 z m−1 + .. + bm a0 z n + a1 z n−1 + .. + an m−n z = b0 z m + b1 zm − 1 + .. + bm Case 1: Consider the case m > n. R(z) has zero of order m − n at ∞ and no poles at ∞ and number of zeros = n and hence the total number of zeros = m − n + n = m. 19

The number of poles of R(z) = m Therefore, number of poles = number of zeros = m. Case 2: m
an bm

an bm

6= 0 since an 6= 0, bm 6= 0

6= ∞ since bm 6= 0.

∴ Number of zeros of R(z) = n = P and Number of poles of R(z) = m = P . The counts show that total of zeros and poles including those at ∞ are always =order of R(z) = P . Note: A rational function of order 1 is a linear function. which is S(z) =

az + b cz + d

with

ad − bc 6= 0

Such fraction/transformation is called linear fractional/Bilinear/mobius transformation . We note w = s(z) has exactly one root. z = s−1 (w) az + b w = cz + d ⇒ czw + dw = az + b z(cw − a) = b − dw dw − b z = − cw − a The transformation s and s− 1 are inverse to each other. The transformation z + a is called parallel translation and

1 z

is an inversion. 20

Result: Every rational function has a partial fraction expansion. Proof. Let R(z) be rational function. R(z) =

P (z) . Q(z)

If degP (z) > degQ(z) that is

n>m ⇒ R(z) has a pole at ∞. By division algorithm, P (z) = Q(z)g(z) + r(z) In other words R(z) = G(z) + H(z)

where

G(z) = g(z)

and

H(z) =

r(z) . Q(z)

Here G(z) is a polynomial without constant term and H(z) is a rational function with property that deg(denominator) ≥ deg(numerator) Suppose deg(P (z)) ≤ deg(Q(z)) ⇒ R(z) = H(z). The above condition on H(z) is same as H(z) =

r(z) . Q(z)

H(z) is finite at ∞ .

The deg(G(z)) is same as the order of pole at ∞ for R(z)(Since G(∞) = ∞) and the polynomial G(z) is also called singular part of R(z) at z = ∞. Summarizing we get if R(z) has a pole at z = ∞, then R(z) = G(z) + H(z) where G(z) is without constant term and H(z) is finite at z = ∞. Let us denote finite distinct poles of R(z) as β1 , β2 , . . . , βl . Consider the new function R(βj + ζ1 ) as a rational function of ζ. Clearly it has a pole at ζ = ∞. Applying the previous case , we can write 1 R(βj + ) = Gj (ζ) + Hj (ζ) ζ

(1.22)

where Gj (ζ) is a polynomial without constant term . Hj (ζ) is finite at ζ = ∞. Put z = βj +

1 ζ

21

Then equation (1.22) ⇒ R(z) = Gj (

1 1 ) + Hj ( ) z − βj z − βj

where   1 1 Gj z−βj is a polynomial in z−β without constant term(singular part of R(z) at j   1 z = βj ) and Hj z−βj is finite at z = βj . This is true for all βj for 1 ≤ j ≤ l. Now consider a new expression, l X

S(z) = R(z) − G(z) −

 Gj

j=1

1 z − βj



This is a rational function in z. The possible poles for R(z) are β1 , β2 , . . . , βl and 1 ∞. G(z) has no poles except at ∞ and Gj ( z−β ) has no poles except β1 , β2 , . . . , βj . j

Thus the only poles for S(z) are β1 , β2 , . . . , βl and ∞. At z = ∞, both R(z) and G(z) become ∞ but their differences H(z) is finite at z = ∞.   1 Similarly at each βj , R(z) and corresponding Gj z−βj become ∞ but there   1 differences Hj z−βj is finite at z = βj . Thus S(z) has neither finite poles nor a pole at z = ∞. Therefore the order of rational function S(z) = 0. ∴ S(z) is constant and absorbing this constant in G(z), then R(z) = G(z) +

l X

 Gj

j=1

1 z − βj



This is a partial fraction. Arc and closed curve Definition: 4 (Curve). We know that the equations of the type x = x(t), y = y(t) give parametric representation of curve in the plane. Using complex variables, the above Equation can be written as z = z(t) = x(t) + iy(t)

where 22

z = x + iy.

Definition: 5 (Arc). The equation of arc γ in the plane is most conveniently given in parametric form x = x(t), y = y(t) where t runs through interval α ≤ t ≤ β and x(t), y(t) are continuous functions. It is also identified that γ is a continuous function of [α, β]. It is denoted by z = γ(t). We call z(α), z(β) the end points of γ. z(α) is the initial point ; z(β) is the terminal point. Definition: 6 (Closed arc/curve). If z(α) = z(β), the curve γ is a closed arc or curve . Definition: 7 (Jordan curve). A curve γ is a Jordan arc simple curve if there exist some parametric representation z = z(t) = x(t) + iy(t), where α ≤ t ≤ β 3 t1 ‘ 6= t2 ⇒ z(t1 ) 6= z(t2 ). (i.e) z(t) is one to one. Definition: 8 (Simple closed curve). A closed curve γ is called simple if there exist a parametric representation z = z(t) = x(t) + iy(t) where α ≤ t ≤ β such that t1 < t2 and z(t1 ) = z(t2 ) implies t1 = α and t2 = β. We usually refer to such curves as simple closed Jordan curves. Note: (i). A closed curve is simple if no two points of it coincide except at the end points. Example 1 Then curve γ1 is defined by z(t) = t2 where −1 ≤ t ≤ 1 is closed but not simple. It is closed since z(−1) = z(1) = 1 since t1 : α 6=

−1 2

−1 2

<

1 2

: t2 but z( −1 ) = z( 21 ) but 2

and β = 21 . Therefore it is not simple.

Example 2. The circle z = cos(t) + i sin(t), 0 ≤ t ≤ 2π is both closed and simple, since value z(t) coincide only at the endpoints t = 0 and t = 2π. Definition: 9 (Opposite arc). The opposite arc of z = z(t) where α ≤ t ≤ β is the arc z = z(−t) where −β ≤ t ≤ −α. They are sometimes denoted by −γ point curve. Definition: 10 (Point curve). A constant function z(t) defined a point curve. 23

Note: (*) A circle C defied as a locus |z − a| = r can be considered as a closed curve with the equation z = a + reiθ , 0 ≤ θ ≤ 2π Definition: 11 (Analytic Functions in Regions). A complex valued function f (z) defined on a open set Ω is said to be analytic in Ω if it has a derivative at each point of Ω. Note: • Every region is a open connected set. • Let f be a complex valued function of complex variable defined in a region Ω, we say that f (z) has a derivative at z in Ω if f (z + h − f (z) h→0 h lim

exists finitely. We call it as f 0 (z). we say that f is Holomorphic /Analytic at Ω if f 0 (z) exist at every z ∈ Ω. Since we have defined analyticity only in regions, we agree that whenever we say f is analytic or Holomorphic in a arbitrary subset A of complex plane we mean f is analytic in the region containing A. At z0 , f is analytic if and only if f is analytic in a disc containing z0 . Definition: 12 (Restriction). A function f (z) is analytic on an arbitrary point set A if it is the restriction to A of a function which is analytic in some open set containing A. Definition: 13 (Single-valued Function). ω is said to be single-valued if one value of z corresponds to only one of ω. Definition: 14 (Multivalued Function). ω is multi-valued if one value of z corresponds to more than 1 value of ω. 24

Example:

√

√ z, n z, log(z), z α , sin−1 z, cos−1 z are multivalued.

Note: Consider the region Ω, the complement of negative real axis z ≤ 0 (i.e., Ω is open √ connected set). In Ω, one and only one of the values of z has a positive real part. √ √ With this choice z is a single valued function in Ω. To prove that w = z is continuous. Given z1 , z2 ∈ Ω and denote the corresponding values of ω by ω1 = u1 + iv1 , ω2 = u2 + iv2 with u1 , u2 ≥ 0, then |z1 − z2 | = |ω12 − ω22 | = |ω1 − ω2 ||ω1 + ω2 | where |ω1 + ω2 | = |(u1 + u2 ) + i(v1 + v2 )| =

p (u1 + u2 )2 + (v1 + v2 )2

≥ u1 + u2 > u1 ∴ |z1 − z2 | > u1 |ω1 − ω2 | |z1 − z2 δ |ω1 − ω2 | < = = u1 u1 Given  < 0 there exist δ = u1 > 0 such that |ω1 − ω2 | < . whenever |z1 − z2 | < δ √ ∴ ω = z is continuous at z1 , To prove differentiability: ω 2 = z ∆ω dω 1 1 1 ∆ω = lim = = dz = = √ ∆z→0 ∆z ∆ω→0 ∆z dz 2ω 2 z dω lim

The point z=0 is a type of singularity called a Branch point. ω = log z: Analytic branch: An analytic branch of ω = log z is given by Ω where Ω:complement of negative real axis in complex plane.We know ∀z ∈ Ω with −π < arg(z) < π ∃ a unique log z with −π < Im(log z) < π.This branch of log z is usually represented as the principal branch of log z. 25

Proof. We first observe that ∀z ∈ Ω ∃ a unique argument θ with |θ| < π with this value for arg(z),we write z = reiθ and infinite values of log z are ω = log z = log(reiθ ) k = 0, ±1, ±2, ...

= log r + i(2kπ + θ), For k = 0,we get one log(z) = log r + iθ with |Im(log z)| = |θ| < π and for all other values of log z, we get |Im(log z)| = |2kπ + θ|,

k = ±1, ±2, ...

Thus for these values |Im(log z)| = |2kπ − (θ)| ≥ |2kπ| − |(θ)|

(Since |θ| < π ⇒ −|θ ≥ −π|, |Im(log z)| ≥ π)

|Im(log z)| ≥ 2|k|π − π |Im(log z)| > π Thus there exists only one value for z with |Im(log z)| < π.Thus ω = f (z) = log z with |Im(logz)| < π becomes one-one map of Ω into {ω|Im(log z)| < π}. This map is also onto.The set {ω/|Im(ω)| < π} as can be seen by taking ω0 with |Im(ω0 )| < π and considering z0 = eω0 . Now ω0 is one of infinitely many values of log z0 and |Im(ω0 )| < π.We see that f (z0 ) = ω0 . Thus ω = f (z) gives one-one correspondence between Ω and {ω/|Im(ω)| < π} To show that : The branch ω = f (z) = log z with |Im(log z)| < π is continuous in Ω.

The principal branch of log z can be written as f (z) = logz = logr + iθ = log|z| + iarg(z) 26

with

|arg(z)| < π

The above definition is valid in Ω. From Real analysis, x → log x (ie) (0, ∞) → R is continuous and onto. Further z → |z| is continuous. ∴ log |z| = Re(f (z)) is continuous in Ω. The Im(f (z)) represent the principal branch of arg(z) and as such it is continuous. ∴ log |z| + iarg(z) is continuous⇒ f (z) is continuous. The continuity of principal branch of ω = f (z) = log z in Ω shows that if z → z0 ,then ω → ω0 .Thus  lim

z→z0

w − w0 z − z0

−1 z − z0 lim z→z0 w − w0  w −1 e − ew0 lim w→w0 w − w0   w − w0 lim w→w0 ew − ew0 1 1 1 lim w = w0 = e−w0 = w→w0 e e z0 

 = = = =

Therefore f (z) is analytic. Analytic branch:w = cos−1 z The infinitely many values of arc cos z are same as values of i log(z +

√ z 2 − 1) in

this case we restrict z to complement Ω of the half lines x ≤ −1, y = 0 and y = 0 0

x ≥ −1 and y = 0. Since 1 − z 2 is never and ≤ 0 in Ω we can define √ √ √ 1 − z 2 as in first example and set z 2 − 1 = i 1 − z 2 . Moreover + z 2 − 1 is √ √ never real in Ω. For z − z 2 − 1 and z + z 2 − 1 are reciprocal and real only if z √ z 2 − 1 are both real. This happens only when z lie on the excluded part of the R √ axis. Because Ω is connected. It follows that all values of z + z 2 − 1 in Ω are on the same side of the real axis and since i is such a value they are all in the upper half plane. We can define an analytic branch of log(z +

√ z 2 − 1) where imaginary part lies

between 0 and π. In this way we have to find a single valued analytic function

27

arccos z = i log(z +

where

√ z 2 − 1) in Ω whose derivatives is   i 2z √ = 1+ √ z + z2 − 1 2 z2 − 1 √ 2  i z −1+z √ √ = z + z2 − 1 z2 − 1 1 = √ 1 − z2

√ 1 − z 2 is in positive real axis.

Theorem: 2. An analytic function in region Ω whose derivative vanishes identically must reduce to a constant. The same is true if Re, Im, modulus or arg is constant. Proof. Let f = u+iv and such that f is complex valued function of complex variable z with f 0 (z) = 0 using the expression for derivative and C − R equations, f 0 (z) = since

∂u ∂x

∂v ∂u ∂v ∂u +i = +i ≡0 ∂x ∂x ∂y ∂y

= 0 ⇒ is a constant on each horizontal lines parallel to x-axis. It follows

that u and v are constant on any line segment Ω which is parallel to one of the axis. But in Ω any two points can be joined as a polygon whose sides are horizontal and vertical. Thus if z0 ∈ Ω is fixed and z ∈ Ω is arbitrary then f (z0 ) = u0 + iv0 = u + iv = f (z) ⇒ u + iv is constant throughout the region. If u (or) v is constant, f 0 (z) =

∂u ∂v ∂v ∂u −i = + =0 ∂x ∂y ∂y ∂x

∴ f (z) is constant. |f (z)|2 = u2 + v 2 If |f (z)|2 (or) u2 + v 2 is constant. ∂u ∂v + 2v = 0 ∂x ∂x ∂u ∂v ⇒u +v = 0 ∂x ∂x 2u

28

And also we get u

∂u ∂v ∂v ∂u +v = 0 ⇒ −u +v =0 ∂y ∂y ∂x ∂x ∂u ∂v ∴ = =0 ∂x ∂x

∴ u and v are constant and similarly ∂u ∂v = =0 ∂y ∂y |f (z)| is constant. Hence Re and Im parts are constant. If arg(f (z)) is constant we can set u = kv, k-constant 6= 0 ∴ f (z) = kv + iv = v(k + i)   1 −1 −1 v arg[f (z)] = tan ( ) = tan kv k But u − kv = 0 is the real part of (1 + ik)f (z) = u − kv + i(uk + v) and we conclude that f(z) must reduce to a constant. CONFORMAL MAPPING: Suppose that an arc γ with equation z = z(t) = x(t) + iy(t) where α < t < β is contained in a region Ω and let f (z) be defined and continuous in Ω, then ω = ω(t) = f (z(t)) define an arc γ 0 in ω-plane; image of γ.Consider the function f (z) which is analytic in Ω,z 0 (t) exist ⇒ ω 0 (t) exist and ω 0 (t) = f 0 (z(t)).z 0 (t) If the mapping of Ω by ω-plane ω = f (z) is topological, then the inverse function f −1 (ω) = z is also analytical this follows that if f 0 (z) 6= 0 then derivative of inverse function must be equal to

1 f 0 (z)

at the point z = f −1 (ω).

The knowledge that f 0 (z) 6= 0 is sufficient to conclude that the mapping is topological, if it is restricted to a sufficiently small neighbourhood of z0 . By Implicit function theorem, the Jacobian of function u = u(x, y) and v = v(x, y) at z0 . J |z0 =

∂u ∂x ∂v ∂x

∂u ∂y ∂u ∂x

= f 0 (z0 )|2 6= 0

29

Definition: 15 (Length and Area). We have found that under a conformal mapping f (z) the length of infinitesimal line segment at the point z is multiplied by |f 0 (z)|.We know from calculus that length of differentiable arc γ with equation z = z(t) = x(t) + iy(t) where α ≤ t ≤ β is given by Zβ p Zβ L(γ) = x0 (t)2 + y 0 (t)2 dt = |z 0 (t)|dt α

α

Definition: 16 (Differentiable arc). If the derivative z 0 (t) = x0 (t) + y 0 (t) exist and not equal to zero the arc γ has a tangent whose direction is determined by arg(z 0 (t)).We say that the arc is differentiable if z 0 (t) exist and is continuous. Definition: 17 (Image curve). The image curve γ 0 is determined by ω = ω(t) = f (z(t)) with derivative ω 0 (t) = f 0 (z(t)) ∗ z 0 (t) its length is L(γ 0 ) =

Zβ

|f 0 (z(t))||z 0 (t)|dt

α

In short notation we denote, L(γ) =

R

|dz| and L(γ 0 ) =

R

|f 0 (z)||dz 0 |

γ0

γ

We observe that in complex notation, ds is replaced by |dz|. Let E be point set RR in the plane where area dxdy can be evaluated as a double Riemann-integral. If f (z) = u(x, y) + iv(x, y) to be a bijective differentiable mapping then by the rule of changing integration variables the area of image E 0 = f (E) is given by Z Z Z Z 0 A(E ) = |(ux vy − vx uy )|dxdy = |f 0 (z)|2 dxdy E0

E0

LINEAR TRANSFORMATION The transformation T is defined by ω = T (z) =

az + d cz + d

(1.23)

where a, b, c, d are complex constant where ad − bc 6= 0 is a bilinear/linear fractional/Mobius transformation the constant ad − bc is the determinant of transformation. (1.23) is normalised if ad − bc = 1, the transformation may be written 30

as, czω + dω − az − b = 0

(1.24)

It is evident that (1.24 ) is linear both in z and ω. The inverse transformation T −1 is defined by z = T −1 (ω) =

−dω + b cω − a

(1.25)

The determinant −d(−a) − bc = ad − bc 6= 0 [same as T ]. NOTE: The transformation T associates a unique point of ω plane to each point of zplane except z =

−b c

then c 6= 0. Also the inverse transformation T −1 associate a

unique point of z-plane to each point of ω-plane except the point ω = These exceptional points z =

−b c

and ω =

a c

a c

when c 6= 0.

are mapped into the point ω = ∞ and

z = ∞ respectively.If the complex plane is closed by addition of point ∞, then we can say that every bilinear transformation sets up a 1-1 correspondence between all points of closed z-plane and closed ω-plane. Remark: Note that the condition ad − bc 6= 0 is essential for bilinear transformation T to set up a 1-1 correspondence between points on closed z-plane and closed ω-plane. If ad − bc = 0,then R.H.S of (1.23)is either a constant or meaningless. Let ω1 and ω2 be the points corresponding to z1 and z2 given by (1.23) as ω1 = ω2 =

az1 +b cz1 +d

az2 +b .Then cz2 +d

ω2 − ω1 = = = = =

az2 + b az1 + b − cz2 + d cz1 + d (az2 + b)(cz1 + d) − (az1 + b)(cz2 + d) (cz2 + d)(cz1 + d) acz1 z2 + daz2 + bcz1 + bd − acz1 z2 − adz1 − bcz2 − bd c2 z1 z2 + cz2 d + dcz1 + d2 ad(z2 − z1 ) + bc(z1 − z2 ) (cz2 + d)(cz1 + d) (ad − bc)(z2 − z1 ) (cz1 + d)(cz2 + d) 31

and

If ad − bc 6= o ⇒ ω2 − ω1 = 0 ⇒ ω2 = ω1 It follows that ω is constant if ad − bc = 0 provided z1 6=

−d c

(or)z2 6=

−d c

Then ω is meaningless.

critical point: az+b cz+b

We notice that the transformation ω =

set up one to one correspondance

between points of closed z-plane and closed ω-plane. Differentiate with respect to z dω (z + d)a − (az + b)c = dz (cz + d)2 acz + da − azc − bc ad − bc = = 2 (cz + d) (cz + d)2 If z = −d/c ⇒ Then

dω dz

= ∞. Ifz = ∞ ⇒ Then

dω dz

=0

The points z = −d/c and z = ∞ are the critical points where the conformal property does not hold and these are the only critical points. We say that a transformation maps a neighborhood of point z0 conformally into a neighborhood of ω = ∞ If the transformation is r =

1 f (z)

maps the neighborhood

of z0 conformally into a neighborhood of r = 0 Again ω = f (z) is said to transform the neighborhood of z = ∞ conformally of ω0 provided ω = φ(x) = f ( τ1 ) transforms the neighborhood of τ = 0 conformally into a neighborhood of ω0 This states that ”Every bilinear transform represents a one to one conformal mapping of whole closed z-plane onto whole w-plane ”. This is called biuniform transformation. Remark: since a bilinear transformation T (z) =

az+b cz+d

represents a conformal mapping, it

preserves angles both in magnitude and sense. Resultant/product of two Bilinear Transformation: Consider transformations T1 and T2 defined by ζ = T1 (z) =

a)1 z + b1 , a1 d1 − b1 c1 6= 0 c1 + d 1 32

(1.26)

and a2 r + b2 , a2 d2 − b2 c2 6= 0 (1.27) c2 r + d 2 (1.26) setsup a one to one correspondence between the points of z-plane and points ω = T2 (r) =

of ζ-plane. (1.27) setup one to one correspondence between points of ζ-plane and point of w-plane. We may define a transformation from z-plane to w-plane by the relation w = T2 (T1 (z))

(1.28)



 a1 z + b1 w = T2 c z + d1 1  a1 z+b1 a2 c1 z+d1 + b2   = a1 z+b1 c2 c1 z+d1 + d2 a2 (a1 z + b1 ) + b2 (c1 z + d1 ) c2 (a1 z + b1 ) + d2 (c1 z + d1 ) z(a2 a1 + b2 c1 ) + a2 b1 + b2 d1 = z(a1 c2 + d2 c1 )b1 c2 + d2 d1 αz + β = γz + δ

=

where α = a1 a2 + b2 c1 ; β = a2 b1 + b2 d1 ; γ = a1 c2 + d2 c1 ; δ = b1 c2 + d2 d1 To find αγ − βγ αγ − βγ = (a1 a2 + b2 c1 )(a1 c2 + d2 c1 ) − (a2 b1 + b2 d1 )(b1 c2 + d2 d1 ) = a21 a2 c2 + a1 d2 c1 a2 + b2 a1 c1 c2 + b2 d2 c21 − a2 b21 c2 − a2 d2 b1 d1 −b2 b1 d1 c2 − b2 d2 d21 = (a21 − b21 )a2 c2 + (c21 − d21 )b2 d2 − (a2 d2 + b2 c2 )b1 d1 + a1 c1 (a2 d2 + b2 c2 ) = (a2 d2 + b2 c2 )(a1 c1 − b1 d1 ) Hence (1.28) is a bilinear transformation. The transformation T2 T1 defined by (1.28) is resultant of T1 and T2 in that order. Hence the resultant of finite number of transformation is also bilinear. 33

Example: Consider w = T1 (z) =

z+2 , z+3

w = T2 (z) =

z z+1

find (i)T1−1 (ii)T2−1 (iii) T2 T1 (z) (iv) T1 T2 (z) (v) T2−1 T1 (z) (i)T1−1 (z)

z+2 ⇒ wz + 3w = z + 2 z+3 wz − z = 2 − 3w 2 − 3w z = = T1−1 (w) w−1 ⇒w =

(ii)T2−1 (w) z ⇒ zw + w = z z+1 z(w − 1) = −w −w z = = T2−1 (w) w−1 ⇒w =

(iii) T2 T1 (z)  ⇒ T2

z+2 z+3



 = =

z+2 z+3 z+2 + z+3

 1

z+2 2z + 5

(iv) T1 T2 (z) = T1 (

z )= z+1

z z+1 z z+1

+2 3z + 2 = +3 4z + 3

(v)T2−1 T1 (z) = T2−1 (

− z+2 z+2 ) = z+2z+3 = z + 2 z+3 −1 z+3

(vi) T1−1 T2 (z) = T1−1 (

z 2z + 2 − 3z )= =z−2 z+1 z−z−1

Remark 3. Consider the bilinear transformation w = 34

az+b cz+d

with ad − bc 6= 0

Case 1: If c 6= 0, then it can be written as (bc − ad) a + c2 (z + dc ) c bc2 − adc + dc(z + ac = c2 (z + dc ) bc2 − adc + dcz + ad = cz − dc

w =

Let us take d Z = z+ , c 1 ζ = , Z bc − ad τ = c2 a ∴ w = τζ + c

(1.29) (1.30) (1.31) (1.32)

These show that a bilinear transformation is resultant of translation in (1.29), inversion in real axis and unit circle in (1.30), rotation and magnification in (1.31), and again translation in (1.32). Case 2: If c = 0 ⇒ w =

az+b d

= ad z + db . Writing ζ = ad z ⇒ w = ζz + db It is resultant

rotation and magnification and translation. Note: Set of all bilinear transformations form a group and also non abelian under resultant of bilinear transformation. Four of our elementary transformations are • Translation • Rotation • Inversion • Homothetic They transform straight line onto straight lines and circles onto circles since they are linear. 35

Suppose we take a translation w = z + a. If z varies over the circle |z − α| = r then w varies over the circle |w − (a + α)| = r If z varies over the straight line z = α + βt, then w varies over the straight line w = a + α + βt.

However inversion does not always take circles onto circles or staright lines onto straight lines. under inversion straight line can be mapped onto a straight or circle .similarly a circle is mapped onto a straight line or circle. consider, The equation αzz + βz + βz + γ = 0

(1.33)

where α, γ -real, β-complex with β β¯ > αγ. If α 6= 0 (1.33) represent a circle. if α = 0 (1.33) is straight line. We find the locus of z1 , from above equation w=

1 1 ⇒z= z w

1 1 1 ¯ 1)+γ = 0 (1.33) ⇒ α( )( ) + β( ) + β( w w¯ w w¯ ¯ + γww¯ = 0 α + β w¯ + βw

(1.34)

Ifγ 6= 0 then (1.34) is a circle. If γ = 0 then (1.34) is straight line thus an inversion maps a circle through origin onto a straight line and maps circle not through origin (γ 6= 0) onto the circles. In a similar way, inversion takes straight lines not through origin (γ 6= 0)onto circles. Remark 4. Any bilinear transformation is a composition of elementary transformation that takes family of circles and straight lines onto straight lines. cross ratio: The cross ratio of four point z1 , z2 , z3 , z4 in extended complex plane where z2 , z3 , z4 are distinct and defined as the image of z1 under bilinear transformation that takes z2 , z3 , z4 into 1, 0, ∞ in that order and denoted by (z1 , z2 , z3 , z4 ) Moreover if none of the point is ∞, then we have sz =

(z − z3 )(z2 − z4 ) (z − z4 )(z2 − z3 ) 36

If z2 or z3 or z4 any one is ∞ z1 − z3 ifz2 → ∞; z1 − z4 z1 − z3 ifz4 → ∞; (z1 , z2 , z3 , z4 ) = z2 − z3 z2 − z4 ifz3 → ∞; (z1 , z2 , z3 , z4 ) = z1 − z4 (z1 , z2 , z3 , z4 ) =

since sz is uniquely determined by these conditions, the cross ratio is well defined. Note: The cross ratio of four points z1 , z2 , z3 , z4 in this order is defined to be equal to z1 −z3 z1 −z4 z2 −z3 z2 −z4

=

(z1 − z3 )(z2 − z4 ) (z1 − z4 )(z2 − z3 )

Remark 5. If T is an another bilinear transformation with the same property, then sT −1 could leave 1, 0, ∞ invariant sz =

(z−z3 )(z2 −z4 ) (z−z4 )(z2 −z3 )

where sz2 = 1, sz3 = 0, sz4 = ∞

Definition: 18 (cross ratio). The cross ratio z1 , z2 , z3 , z4 ) is the image of z1 under bilinear transformation which comes z2 , z3 , z4 into 1, 0, ∞ Theorem: 3. If z1 , z2 , z3 , z4 are distinct points in extended plane and T is any bilinear transformation, then the cross ratio of (Tz1 , Tz2 , Tz3 , Tz4 ) = (z1 , z2 , z3 , z4 ) Proof. If s is a bilinear transformation, then it takes is z2 , z3 and z4 to 1, 0, ∞ Then by definition of cross ratio.

sz =

(z − z3 )(z2 − z4 ) = (z1 , z2 , z3 , z4 ) (z − z4 )(z2 − z3 )

we know that sT −1 also carries z2 , z3 , z4 into 1, 0, ∞ (Tz1 , Tz2 , Tz3 , Tz4 ) = sT −1 (Tz1 , Tz2 , Tz3 , Tz4 ) = (sz1 , sz2 , sz3 , sz4 ) = (z1 , z2 , z3 , z4 )

37

Theorem: 4. The cross ratio (z1 , z2 , z3 , z4 ) is real if and only if the four points lie on a circle or a straight line. Proof. We know that image of R under bilinear transformation is either a circle or straight line. If the cross ratio of four points (z1 , z2 , z3 , z4 ) is real where sz2 = 1, sz3 = 0, sz4 = ∞ Therefore z2 = s−1 (1); z3 = s−1 (0); z4 = s−1 (∞); so z1 , z2 , z3 , z4 lie on the image of real axis under s−1 which by our assumption is either a circle or a straight line. On the other hand if cross ratio of four points are real, then points lie on circle or a straight line. Otherwise if four points z1 , z2 , z3 , z4 lie on a straight line on a complex plane.Take s such that sz2 = 1, sz3 = 0, sz4 = ∞ as s−1 takes the real line onto a circle or a straight line and in as much as z1 , z2 , z3 , z4 belongs to this image C = s−1 (R) R = S(C) For all complex z ∈ C, Sz is real. We have to proved that the image of real axis under any bilinear transformation is either a circle or a straight line. Let T be a bilinear transformation T −1 (w) = z =

aw+b cw+d

We find the locus of w when z satisfies z = z¯ then aw + b a ¯w¯ + ¯b = cw + d c¯w¯ + d¯ c¯aww¯ + awd¯ + b¯ cw¯ + bd¯ = a ¯cww¯ + cw¯b + d¯ aw¯ + ¯bd ¯ − ¯bc) + w(¯ ¯ − ¯bd) = 0 ww(¯ ¯ ca − a ¯c) + w(da ¯ cb − a ¯d) + (db

(1.35)

¯ − ¯bd are purely imaginary then multiplying (1.35) by i, Note that c¯a − a ¯c and db ¯ − ¯bc) + iw(¯ ¯ − ¯bd) = 0 iww(¯ ¯ ca − a ¯c) + iw(da ¯ cb − a ¯d) + i(db αww¯ + βw + β¯w¯ + γ = 0 38

(1.36)

where α = i(a¯ c − c¯ a) , β = i(ad¯ − c¯b) , β¯ = i(¯bc − d¯ a) , γ = i(bd¯ − ¯bd) we observe that (1.36 ) conforms the standard equation of circle or a straight line. It is a circle, if (a¯ c − c¯ a) 6= 0 (i,e)., α 6= 0 Straight line, if (a¯ c − c¯ a) = 0 (¯bc − d¯ a), (ad¯ − c¯b) 6= 0 (i.e) α = 0, β, β¯ = 6 0. Theorem: 5. The bilinear transformation transform circles onto a circle and straight line onto a straight lines.

39

Chapter 2

COMPLEX INTEGRATION Introduction In real variable, a distinction is made between definite and indefinite integrals. The former being regarded as limit of sum and latter as process inverse to differentiation. We make a similar distinction between definite and indefinite integrals of complex variables. Definite integrals of complex variables are usually known as line integrals. As in the case of real variable, as indefinite integral of a complex variable is a complex function whose derivatives equals a given analytic function in a region. The indefinite integrals of any elementary function can be obtained by inversion of known derivative formula. However, the theory of definite variable does not extend their straight way to complex variable domain. Example: In the case of R variable, the path of integration of

Rb

f (x)dx is always

a

along R from x = a to x = b. In the case of complex function f (z), the path of definite integral may be along any curve joining z = a to z = b. Thus generally value of integral depends upon the path of integration. Line integrals The most immediate generalization of real integral is to the definite integral of complex function over a real interval.

40

If f (t) = u(t) + iv(t) is a continuous function defined in(a, b), we set by definition, Zb

Zb f (t)dt =

a

Zb u(t)dt + i

a

v(t)dt

(2.1)

a

This integral has most of the properties of real integral. If c = α + iβ is a complex constant, Zb

Zb cf (t)dt = c

a

f (t)dt

(2.2)

a

for both members are Zb

Zb

(α + iβ)(u(t) + iv(t))dt

cf (t)dt = a

a

Zb

Zb u(t)dt − β

= α a

Zb v(t)dt + i

a

[αv(t) + βu(t)]dt a

If a ≤ b, we know that Zb Zb f (t)dt ≤ f (t) dt a

(2.3)

a

holds for arbitrary function. If we choose c = e−iθ , where θ is real, equation (2.2) becomes  b  Z Zb   −iθ Re  e f (t)dt = Re e−iθ f (t) dt a

a

Zb ≤

|e−iθ f (t)|dt

a

Zb |f (t)|dt

= a

 b  Z Zb −iθ Re  e f (t)dt ≤ |f (t)|dt a

a

41

For θ = arg

Rb

|f (t)|dt, left hand side of the equation reduces to absolute value of

a

the integral.

Definition: 19 (Piece-wise Differentiable Arc). An arc γ is said to be piece-wise diffferentiable if there is an equation z = z(t) = x(t)+iy(t), where t runs through a ≤ t ≤ b and x(t) and y(t) are continuous functions of t in complex plane, there exists [a, b] can be divided into finite number of sub intervals, [a, t1 ], [t1 , t2 ], . . . , [tn−1 , b] on each of which z 0 (t) exist and in addition z 0 (t) 6= 0 in any of these interval, then γ is piece-wise differentiable or regular or smooth arc. If f (z) is defined and continuous on γ, then f (z(t)) is also continuous and we can set R

f (z)dz =

γ

R

Z

Z f (z)dz =

γ

f (z(t))dz(t)

γ

f (z(t))z 0 (t)dt

(2.4)

γ

This is the definition of complex line integral of f (z) over an arc γ. In right hand side of equation (2.4), if z 0 (t) exist and not continuous throughout the interval of integration, then the interval should be subdivided into obvious manner. Whenever a line integral over an arc γ is considered ,it is understood that γ is piece-wise differentiable. Property: The important property of equation (2.4) is its invariance under change of parameter which is determined by an increasing function t = t(τ ) which maps an interval α ≤ τ ≤ β onto a ≤ t ≤ b. We assume t(τ ) is piece-wise differentiable . By the rule for changing the variable of integration we have Rb Rβ f (z(t))z 0 (t)dt = f [z(t(τ ))]z 0 (t(τ ))t0 (τ )dτ a

α

But z 0 (t(τ ))t0 (τ ) is the derivative of z(t(τ )) with respect to τ and hence equation (2.4) has the same value whether γ be represented by the equation z = z(t) or 42

z = z(t(τ )) Integral of opposite arc: Consider an opposite arc z = z(−t) where −b ≤ t ≤ −a we have Z−b

Z f (z)dz = −γ

f (z(−t))z 0 (t)(−dt)

−a

Z−b = −

f (z(−t))z 0 (t)(dt)

−a

And by a change of variable s = −t,ds = −dt Zb

Z f (z)dz = −γ

f (z(s))z 0 (s)ds

a

Zb = −

f (z(s))z 0 (s)ds

Za

Z f (z)dz = − −γ

f (z)dz γ

It is clear that by subdividing an arc γ into a finite number of sub arcs,a subdivision can be represented by γ = γ1 + γ2 + . . . + γn R R R and the corresponding integrals satisfy f (z)dz = f (z)dz + f (z)dz + . . . + γ γ1 γ2 R f (z)dz. γn

Finally, the integral over closed curve is also invariant under shift of parameters R R That is, f (z)dz = f (z)dz. γ1 +γ2

γ2 +γ1

Next we consider the line integrals of z¯, the most convenient definition is by double conjugation R

f (z)d¯ z=

γ

R γ

43

f¯dz

and also we have   Z Z 1 f dx = f dz + f d¯ z 2 γ γ γ   Z Z Z 1  f dz − f d¯ z f dy = 2i

Z

γ

γ

γ

Then equation (2.4) can be written as when f = u + iv R

R f (z)dz = (u + iv)(dx + idy)

γ

γ

Z

Z udx − vdy + i

f (z)dz = γ

Z

γ

By defining integrals of the form

udy + vdx

(2.5)

γ

R

pdx + qdy in which equation (2.5) would serve

γ

as a definition of equation (2.4). A different line integral is obtained by integration with respect to arc length. Two notations are in common use and definition is Z Z Z f ds = f (z)|dz| = f (z(t))|z 0 (t)|dt, γ

γ

γ

(2.6)

since t > 0. This integral is independent of choice of parameters R γ

f dz =

R

f |dz| =

R

f |dz|

−γ

γ

R R The inequality | f dz| ≤ |f ||dz| is a consequence of equation (2.3). γ

γ

Example: For f=1, the equation (2.6) reduces to

R

f ds =

γ

R

|dz| which is by definition of the

γ

length of the arc γ. Now computing the length of the circle, from z = z(t) = a + reit

44

where 0 ≤ t ≤ 2π of full circle, we obtain z 0 (t) = ireit Z2π ⇒

Z2π

0

|z (t)|dt = 0

|ireit |dt

0

Z2π =

reit dt

0

= 2πr RECTIFIABLE ARC The length of arc can be defined as the least upper bound of all sums |z(t1 ) − z(t0 )| + |z(t2 ) − z(t1 )| + . . . + |z(tn ) − z(tn−1 )| where a < t0 < t1 < . . . < tn < b. If least upper bound < ∞, then the arc is rectifiable. Note: Piece-wise differentiable arcs are differentiable. Functions of Bounded Variations We see that (i)|x(tk ) − x(tk−1 )| ≤ |z(tk ) − z(tk−1 )| (ii)|y(tk ) − y(tk−1 )| ≤ |z(tk ) − z(tk−1 )| (iii)|z(tk ) − z(tk−1 )| = |x(tk ) + iy(tk ) − x(tk−1 ) − iy(tk−1 )| ≤ |x(tk ) − x(tk−1 )| + i|y(tk ) − y(tk−1 )| where 0 ≤ k ≤ n. It follows that the sum |z(t1 ) − z(t0 )| + |z(t2 ) − z(t1 )| + . . . + |z(tn ) − z(tn−1 )| and the sum |x(t1 ) − x(t0 )| + |x(t2 ) − x(t1 )| + . . . + |x(tn ) − x(tn−1 )| + |y(t1 ) − y(t0 )| + |y(t2 ) − y(t1 )| + . . . + |y(tn ) − y(tn−1 )| are bounded at the same time. We say that x(t) and y(t) are bounded variations if latter sums are bounded. Result: A necessary and sufficient condition for an arc z = z(t) = x(t) + iy(t) to be rectifiable is that x(t) and y(t) are functions of bounded variations. To show that Rb p Rb arc γ is rectifiable and its length n is given by l = x0 (t)2 + y 0 (t)2 dt = |z 0 (t)|dt = a

45

a

Rb

|dz|.

a

Line integrals as functions of arcs R We have seen that line integral f (z)dz over an arc γ can be put in the form γ R R (u + iv)(dx + idy). General line integrals of the form pdx + qdy are often studied γ

γ

as functions of arc γ under the assumption that p, q are defined and continuous in Ω such that γ is free to vary in Ω. ”There is an important class of integral characterized by the property that the integral over an arc depends only on its end points”. This means that if two arcs γ1 , γ2 having same initial point and same end point, then R

pdx + qdy =

γ1

R

pdx + qdy

γ2

Results: The following are equivalent: (i) A line integral of f (z) over an arc γ depends only on end points of γ. (ii) The integral f (z) over any closed curve is zero. proof: Let the integral f (z) over any closed curve is zero and let γ1 , γ2 be two arcs with R same end points. Then γ1 − γ2 becomes a closed curve. So f (z)dz = 0 which γ1 −γ2

is equivalent to R

R f (z)dz − f (z)dz = 0 γ1 R γ2 R ⇒ f (z)dz = f (z)dz γ1

γ2

Conversely, Let the integral over any two arcs having same end points be equal. Let Γ be any closed curve, then Γ, −Γ have same end points. Then we have R Γ

R f (z)dz = − f (z)dz −ΓR Γ ⇒ 2 f (z)dz = 0 RΓ ⇒ f (z)dz = 0

f (z)dz =

R

Γ

Hence the proof.

46

Theorem: 6. The integral,

R Γ

pdx + qdy defined in Ω if and only if there exists a

function U (x, y) in Ω with partial derivative

∂U ∂x

= p and

∂U ∂y

= q.

Proof. Sufficient Part: Suppose there exist a function U (x, y) in Ω such that ∂U ∂x

= p and

∂U ∂y

= q .Then if a and b are end points γ, we have Z

Z pdx + qdy =

Γ

ZΓ =

∂U dx + ∂x

Z

∂U dy ∂y

Γ

∂U dx ∂U dy .dt + .dt ∂x dt ∂y dt

Γ

Zb  =

 ∂U 0 ∂U 0 x (t) + y (t) dt ∂x ∂y

a

Zb =

d U (x(t), y(t))dt dt

a

= [U (x(t), y(t))]ba = U (x(b), y(b)) − U (x(a), y(a))

(2.7)

Since right hand side of equation (2.7) depends only on a and b. Therefore sufficient part is proved. Necessary part: Let the line integral

R

pdx + qdy depends only on the end points

γ

of γ. We choose a fixed point (x0 , y0 ) in Ω and arbitrary (x, y) in Ω. We join (x0 , y0 ) to (x, y) by a polygonal arc γ contained in Ω whose sides are parallel to coordinate axes. Define a function U by U (x, y) =

R

pdx + qdy

γ

By hypothesis, the integral depends only on the end points and hence it is well defined. If we choose the last segment of γ which is horizontal in which we can keep y constant( dy = 0) and suppose x-varies without changing other segment. Choosing x as parameter on the last segment, we obtain

47

U (x, y) =

Rx

pdx + constant

Z [Note dy = 0 ⇒

qdy = constant].

We do not specify the lower limit since it is immaterial

∂U ∂x

= p.

Similarly choosing last vertical segment, it can be shown that

∂U ∂y

= q.

This proves the necessary condition. Remark 6.

(1) It is customary to write dU =

∂U dx ∂x

+

∂U dy ∂y

and we say dU =

pdx + qdy is an exact differential if it can be written in the above form.” An integral depends only on end points if and only if the integral is an exact differential.” (2) We now determine the condition under which f (z)dz = f (z)dx + if (z)dy is an exact differential. By definition of exact differential, there must exist a function F (z) in Ω such that then

∂F (z) ∂x

∂F (z) ∂x

= f (z) and

∂F (z) ∂y

= if (z) ⇒ −i ∂F∂y(z) = f (z)

= −i ∂F∂y(z) which is a complex form of Cauchy Riemann equations.

Also F (z) is continuous by assumption. Hence F (z) is analytic. (3) From the previous remarks, we conclude that the integral

R

f (z)dz with con-

γ

tinuous f , depends only on the endpoints of γ if and only if f is derivative of analytic function in Ω. Example: For n ≥ 0, the function (z − a)n is the derivative of

(z−a)n+1 n+1

which

is analytic in the whole complex plane. If ζ is any closed curve, then we know that R R (z − a)n dz = 0. If n is negative except n = −1, then (z − a)n dz = 0 for all ζ

ζ

closed curve ζ that does not pass through 0 a0 . Since in the complementary region of point 0 a0 , the indefinite integral is analytic and single valued. If n = −1, then R (z − a)n dz = 0 does not hold. ζ

Consider a circle C with centre a represented by z = a + reit , 0 ≤ t ≤ 2π. We obtain z − a = reit 48

dz = ireit On integrating

R C

dz dz z−a

=

R2π ireit 0

reit

dt = 2πi

Theorem: 7 (CAUCHY THEOREM FOR A RECTANGLE). If the function f (z) R is analytic on a rectangle R then f (z)dz = 0 ∂R

Construction: By a rectangle R in complex plane, we mean a set of points (x, y) such that a ≤ x ≤ b ; c ≤ y ≤ d. We think of parameter of R as a simple closed curve consisting of four line segments whose direction is chosen in such a manner that the area of R lies in the left of these directional segments. We refer this closed curve as boundary curve or contour of R and is denoted by ∂R. The vertices occur in the order (a, c) ; (b, c) ;(b, d) ; (a, d).

Proof. We subdivide the rectangle R into four equal parts and call the individual rectangles R(1) , R(2) , R(3) and R(4) and their boundaries as ∂R(1) , ∂R(2) , ∂R(3) , ∂R(4) . R R Let η(R) = f (z)dz and η(R(k) ) = f (z)dz ∂R (1)

(2)

∂R(k) (3)

⇒ η(R) = η(R ) + η(R ) + η(R ) + η(R(4) ) It follows from above equation that atleast one of rectangles R(k) , k = 1, 2, 3, 4 must satisfy |η(R(k) )| ≥ 41 |η(R)|. We shall fix one such rectangle and call it as R1 so that |η(R1 )| ≥ 41 |η(R)| divide R(1) in the same way and get a particular rectangle R2 such that |η(R2 )| ≥ 41 |η(R1 )| continuing, we get a sequence of rectangle R ⊃ R1 ⊃ R2 ⊃ . . . Rk ⊃ . . . and that |η(Rk )| ≥ 14 |η(Rk−1 )| ≥ . . . ≥

1 |η(R)| 4k

note that diameter of Rk as dk and perimeter by lk satisfying, dk =

d ,l 2k k

=

l 2k

where d and l are diameter and perimeter of R. 49

By cantor intersection theorem, ∞ T Rk is a singleton set say {z ∗ } k=1

and so z∗ belong to every Rk . Given any neighborhood B of z ∗ we can find a large N 3 k ≥ N ⇒ Rk ⊂ B if we choose dk =

d 2k

< ρ for k ≥ N,

then for any z ∈ Rk |z − z ∗ | ≤ dk < ρ so that Rk ⊂ B now z ∗ ∈ Ω and so f (z) is analytic at z ∗ Hence given  > 0 we can find a δ > 0 3 0 < |z − z ∗ | < d, f (z) − f (z ∗ )  0 ∗ ⇒ − f (z ) < ∗ z−z 2dl since f (z ∗ ) and f 0 (z ∗ ) are constant and 1.dz and z.dz are exact differentials

50

R ∂Rk

dz = 0 and

R

zdz = 0

∂Rk

Z η(Rk ) =

f (z)dz ∂Rk

Z =

[f (z) − f (z ∗ ) − f 0 (z ∗ )(z − z ∗ )]dz

∂Rk

Z ⇒ |η(Rk )| ≤

|f (z) − f (z ∗ ) − f 0 (z ∗ )(z − z ∗ )||dz|

∂Rk

Z <

|z − z ∗ | |dz| 2dl

∂Rk

Z ≤

dk |dz| 2dl

∂Rk

dk  d l lk = 2dl 2dl 2k 2k  1 = 2 4k  |η(R)| ≤ 4k |η(Rk )| ≤ <  2 ≤

Definition: 20. Let f (z) be analytic in a region in Ω except at r ∈ Ω, we say that r is exceptional point to f (z) if lim (z − r)f (z) = 0

z→0

Theorem: 8 ( Cauchy’s Theorem for rectangle in the presence of exceptional point). Let f (z) be analytic on the set R0 obtained from rectangle R by omitting a finite number of finite points rj . If it is true that Z lim (z − rj )f (z) = 0 ∀j then f (z)dz = 0 z→rj

∂R

Proof. Construction: The rectangle R can be subdivided into smaller rectangles squares in such a way that each exceptional point occurs as the center of one and only one square and that integral over boundary of R equal sum of integrals over 51

boundaries of square are rectangles without loss of generality, we can assume that R is a square containing one and only one exceptional point r which lies at its center. Let r be an exceptional point given  > 0 ∃ δ > 0 3 0|z − r| < δ  8  ⇒ |f (z)||(z − r)| < 8 ⇒ |f (z)(z − r)| <

also this square can be further be subdivided so that for this δ > 0, r lies at the center of the square R0 of diameter d (its side being Z

√δ ) 2

and that

Z f (z)dz =

∂R

f (z)dz ∂R0

if z varies over ∂R0  |f (z)| < 8|z − r| Z Z and f (z)dz ≤ |f (z)||dz| ∂R

∂R0

 < 8

Z

|dz| |z − r|

∂R0

√ Z 2 2 = |dz| 8 δ ∂R0 √ √ 2 24 δ √ = 8 δ 2 Z ⇒ f (z)dz <  ∂R

Theorem: 9 ( CAUCHY’S THEOREM ON DISC). If f (z) is analytic in open disc R ∆ then f (z)dz = 0 for every closed curve γ in ∆. γ

52

Proof. Let O be a center z0 = x0 + iy0 and P be any point z = x + iy inside ∆. We define a function Z F (z) =

f (z)dz

(2.8)

σ

where σ consists of line segment OA from (x0 , y0 ) to (x, y0 ) and vertical segment AP from (x, y0 ) to (x, y) thus (2.8) may be written as Z Z Z f (z)dz + f (z)dz f (z)dz = F (z) = AP

OA

OAP

(2.9)

Note that on OA, y is constant = y0 and x varies from x0 to x so that we may set z = t + iy0 dz = dt (on OA) similarly on AP , x is constant and y varies from y0 to y and set z = x + it0 dz = idt (on AP ) then (2.9) becomes Zy

Zx F (z) =

f (t + iy0 )dt + i x0

f (x + it)dt

(2.10)

y0

By cauchy’s theorem for rectangle Z f (z)dz = 0 OAP BO

⇒ integral of f (z) over OAP and OBP are same. Accordingly F (z) may also defined by Z

Z

F (z) =

f (z)dz = OP B

Z f (z)dz +

OP

Zy =i

(2.11)

BP

Zx f (x0 + it)dz +

y0

f (t + iy)dz x0

Differentiate (2.10) with respect to y ∂F ∂y

f (z)dz

= if (x + iy) = if (z) 53

(2.12)

Differentiate (2.11) with respect to x ∂F ∂x

= f (x + iy) = f (z)

Hence ∂F ∂F +i ∂x ∂y ∂F ∂x

= f (z) − f (z) = 0 = −i

∂F ∂y

(2.13)

If F (z) = u + iv then (2.7) gives ⇒

∂u ∂x

∂v ∂v + i ∂x + i( ∂u + i ∂y )=0 ∂y

( ∂u − ∂x

∂v ) ∂y

∂v + i( ∂x +

∂u ) ∂y

=0

Equating real and imaginary parts ∂u ∂x

−

∂v ∂y

∂u ∂x

= 0 and

=

∂v ∂y

∂v ∂x

∂v ∂x

Thus u and v satisfy C-R equations since

∂F ∂x

+

∂u ∂y

=0

= − ∂u ∂y = f (z) and

∂F ∂y

= if (z) and F (z) is

continuous. It follows that 4 partial derivatives Ux , Uy , Vx , Vy , are all continuous. Therefore F (z) = u + iv is analytic on ∆. We now find derivative F 0 (z): F 0 (z) =

∂F ∂x

= f (z) and we have proved the following: ” If f (z) is analytic in tthe

disc |z − z0 | < r then there exist another analytic function F (z) in |z − z0 | < r such that F 0 (z) = f (z) ” R It follows that f (z)dz = 0 for any closed curve γ in ∆ γ

Result: Let f (z) be a continuous function defined in a region Ω. The differential f (z)dz is exact if and only if there exist a analytic function U (z) defined on Ω such that U 0 (z) = f (z), that is, f (z) has a primitive in Ω. Proof: Let U (z) be analytic with U 0 (z) = f (z) in Ω. Now f (z)dz = f (z)dx + if (z)dy. Since

54

U (z) is analytic in Ω, we have ∂U = U 0 (z) = Thus U 0 (z) =

∂u ∂x

= f (z) and iU 0 (z) =

Therefore, f (z)dz =

∂u dx ∂x

+

∂u dy ∂y

∂u ∂y

∂u ∂x

= −i ∂u ∂y

= if (z).

= dU and f (z)dz exist.

On the another hand if f (z)dz exist, we have continuous function U (z) defined on Ω such that ∂u ∂x

∂U ∂x

= f (z) and

∂u ∂y

= if (z). But U (z), ∂u , ∂x

∂u ∂y

are continuous and

= −i ∂u . Therefore it satisfies C-R equations. It follows that U (z) is analytic ∂y

with ∂u = U 0 (z) = f (z). Theorem: 10 (CAUCHY THEOREM ON A CIRCULAR DISK WITH EXCEPTIONAL POINTS). Let f (z) be analytic in the region ∆ obtained by omitting a finite number of points rj from anZopen disc ∆. If f (z) satisfies the condition, f (z)dz = 0 for any closed curve γ in ∆. lim (z − rj )f (z) = 0 for all j, then z→rj

γ

Proof. Let Ω be the region consisting of all points of open disc ∆ except a finite number of exceptional points. We show that f has a primitive in Ω. (By previous remark) If we take an analytic function F (z) in Ω satisfying F 0 (z) = f (z), then R f (z)dz = 0 for any closed curve in ∆0. Choose a base point z0 in Ω. For z1 ∈ Ω, γ R we can choose a rectangular path γ joining z0 and z1 . Define g(z1 ) = f (z)dz. If γ

σ is any other rectangular path in Ω joining z0 and z1 , then γ − σ is a rectangular path which is closed in Ω. Then Z Z Z f (z)dz = 0 ⇒ f (z)dz = f (z)dz γ−σ

γ

σ

Thus g(z1 ) is well defined for all z1 ∈ Ω. Since f (z) is continuous at z1 , given  > 0, ∃δ > 0 such that for |z − z1 | < δ implies |f (z) − f (z1 )| < 2 . To choose δ so that the closed δ-neighborhood of z1 does not contain any exceptional point. Put ψ(z) = f (z)−f (z1 ) so that |ψ(z)| <

 2

for |z−z1 | < δ since z1 +h lies in neighborhood,

we have |h| < δ. Having chosen γ joining z0 and z1 , choose the rectangular path σ joining z1 and z1 + h to be γ ∪ σ1 ∪ σ2 where σ1 is the horizontal segment at z1 and R σ2 is the vertical line segment at z1 + h. It is clear that, g(z1 + h) = f (z)dz γ∪σ1 ∪σ2

55

and Z

Z f (z)dz −

g(z1 + h) − g(z1 ) = γ∪σ1 ∪σ2

f (z)dz γ

Z f (z)dz

= σ1 +σ2

and that σ1 and σ2 are fully contained in disc of radius δ at z1 . For every z in this path |z − z1 | ≤ |h| < δ, we have Z g(z1 + h) − g(z1 ) 1 − f (z1 ) ≤ |f (z) − f (z1 )||dz| h |h| σ1 +σ2 Z 1 = |ψ(z)||dz| |h| σ1 +σ2

 |h| <  < 2|h| ⇒ g0(z1 ) = f (z1 ) Since z1 is arbitrary, g 0 (z) = f (z). Hence proved. Cauchy Integral Formula Index of a point with respect to a closed curve: Lemma: If the piecewise differential closed curve γ does not pass through a point R dz ’a’ then the value of integral z−a is a multiple of 2πi. γ

Proof: Here Z

dz = z−a

γ

Z dlog(z − a) γ

Z

Z dlog|z − a| + i

= γ

darg(z − a) γ

When z describes a closed curve log|z − a| returns to its initial value and arg(z − a) increases or decreases by multiple of 2π. 56

Let γ be parametrized by the equation z = z(t) where α ≤ t ≤ β. Let us consider Rt z0 (t) h(t) = z(t)−a dt. h(t) is defined and continuous on [α, β] and has the derivative α

0

h (t) =

z 0 (t) dt z(t)−a

whenever z 0 (t) is continuous. From this equation it follows that

derivative of e−h(t) (z(t) − a) vanishes except at a finite number of points. Now we take F (t) = e−h(t) (z(t) − a) F 0 (t) = e−h(t) [−h0 (t)(z(t) − a)] + e−h(t) z 0 (t) = e−h(t) [−h0 (t)(z(t) − a) − z 0 (t)] = e−h(t) [z 0 (t) − z 0 (t)] = 0 ⇒ F (t) = constant We evaluate the constant at t = α.

∴ eh(t)

= z(α) − a z(t) − a z(t) − a = = c z(α) − a

Since z(α) − z(β) ⇒ eh(β) = 1 Therefore h(β) must be a multiple of 2πi. Index of the point 0 a0 with respect to the closed curve γ: Index of the point 0 a0 with respect to the closed curve γ is given by the equation R dz 1 n(γ, a) = 2πi . z−a γ

Note: The index is also called the winding number of γ with respect to 0 a0 . Property 1: It is clear that Zn(−γ, a) = −n(γ, a)Z f (z)dz f (z)dz 1 1 n(−γ, a) = 2πi =− = −n(γ, a). 2πi γ z − a −γ z − a Property 2: If γ lies inside of a circle, then n(γ, a) = 0 for all points 0 a0 outside the same circle. proof: Let Ω be an unbounded region containing γ. Choose a large circle containing γ in its interior[note: It has a point set of γ and it is unbounded]. We can always 57

choose a point α of γ outside the circle. Using Cauchy theorem for a circular disc,

n(γ, a) = Since

1 z−a

Z

1 2πi

γ

f (z)dz =0 z−a

is analytic at inside of C and γ ⊆ C.

Property3: As a function of 0 a0 , the index of n(γ, a) is constant in each region determined by γ and 0 in the unbounded region. Integral formula Let f (z) be analytic in an open disc 4 and consider a closed curve γ in 4 and a point a in 4 which does not lie in γ. We apply Cauchy theorem to function F (z) =

f (z)−f (a) . z−a

It is analytic for z 6= a.

For z = a, it is not defined but it satisfies lim F (z)(z − a) = 0 that is condition for z→a

Cauchy’s theorem of circular disc with exceptional points. We conclude that Z Z f (z) − f (a) f (z)dz f (a) dz = − dz = 0 z−a Z z − a γ z −a γ Z dz f (z) 1 dz = f (a) ⇒ 2πi γ z −a γ z −a

Z γ

and we observe that

1 2πi

Z γ

dz = f (a)n(γ, a). z−a

Theorem: 11. Suppose that f (z) is analytic in an open Zdisc 4 and let γ be a closed f (z) 1 curve in 4. For any point 0 a0 not in γ, n(γ, a)f (a)= 2πi dz where n(γ, a) is γ z −a theindex of a with respect to γ. Proof. We describe a circle γ defined by |z − a| = ρ with ρ < d where d is the distance of a from γ. Then the function φ(z) =

f (z) z−a

is analytic in the open disc and

γ. Then 1 2πi

R γ

f (z) dz z−a

=

1 2πi

R γ

f (z)−f (a) dz z−a

58

+

1 2πi

R γ

f (a) dz z−a

Writing the circle z − a = ρeiθ ; dz = iρeiθ dθ Z Z f (z) ρieiθ dz = f (a) dθ z−a ρeiθ γ

γ

= f (a)[2πi] and f (z) 1 dz = z−a 2πi

Z

f (z) 1 dz − f (a) = z−a 2πi

Z

1 2πi

Z γ

1 2πi

Z

f (z) − f (a) dz + f (a) z−a

γ

γ

f (z) − f (a) dz z−a

(2.14)

γ

Since f (z) is continuous at a, given  > 0 ∃δ > 0 such that |f (z) − f (a)| <  for all z with |z − a| < δ. Since ρ is at our choice, we can take ρ < δ. So the above inequality is satisfied for all points on γ. Hence

Z Z2π 1 1 f (z) − f (a) f (z) − f (a) iθ = dz iρe dθ 2πi 2πi z−a ρeiθ γ 0

≤

1 2π

Z2π |f (z) − f (a)||dθ| 0

≤

 2π

Z2π |dθ| 0

 ≤ (2πρ) 2π < 

59

From equation (2.14), we get Z 1 f (z) dz − f (a) <  2πi z−a γ Z 1 f (z) dz = f (a) ⇒ 2πi z−a γ Z 1 f (z) n(γ, a)f (a) = dz 2πi z−a γ

It is clear that the above theorem remains valid to which Cauchy theorem for circular disc with exceptional points can be applied in any region Ω. The most common application is to the case n(γ, a) = 1, then we have f (a) =

1 2πi

R γ

f (z) dz. z−a

This we represent as representation formula. It permits us to compute f (a) as soon as values of f (z) on γ are given together with the fact that f (z) is analytic in 4. We let, a → z and z → ζ. 1 f (z) = 2πi

Z

f (ζ) dζ ζ −z

(2.15)

γ

This is called Cauchy Integral Formula. It is valid only when n(γ, z) = 1 and we have proved only when f (z) is analytic in a disc. Higher Derivatives: The representation formula (2.15) gives us an ideal tool for the study of local properties of analytic functions. We can show that, a analytic function has derivatives of all order which are then also analytic. Consider f (z) which is analytic in arbitrary region Ω. To a point a ∈ Ω, we determine a δ-neighbourhood ∆ ⊂ Ω and in ∆, a circle c about a. n(c, a) = 1 for all points z inside c for such z we obtain from (2.15) Z 1 f (ζ) f (z) = dζ 2πi c (ζ − z)2 60

provided that the integral can be differentiated under the sign of integration, we find Z 1 f (ζ) f (z) = dζ 2πi c (ζ − z)2 Z f (ζ) n! n f (z) = dζ 2πi c (ζ − z)n+1 0

Lemma: Suppose φ(ζ) is continuous on arc γ, then the function Z φ(ζ) fn (z) = dζ n γ (ζ − z) is analytic in each of region determined by γ and its derivative fn0 (z) = nfn+1 (z). Proof. We shall prove by induction on n. Take n = 1 and z0 ∈ γ c . Choose δ > 0 such that |z − z0 | < δ, the disc dose not meet γ. Now, restrict z to the bounded |z − z0 | <

δ 2

so that |ζ − z| ≥ δ and |ζ − z0 | = |ζ − z + z − z0 | ≥ |ζ − z| − |z − z0 | δ ≥ δ− 2 δ ∴ |ζ − z0 | ≥ ∀∈γ 2

now Z

Z φ(ζ) φ(ζ) f1 (z) − f2 (z) = dζ − dζ ζ −z γ ζ − z0 γ   Z ζ − z0 − ζ + z ∴ f1 (z) − f1 (z0 ) = φ(ζ)dζ (ζ − z)(ζ − z0 ) γ Z (z − z0 )φ(ζ) = dζ (ζ − z)(ζ − z0 ) γ Z |φ(ζ)|z − z0 | |f1 (z) − f1 (z0 )| ≤ |dζ| |ζ − ||ζ − z0 | γ Z |φ(ζ)| ≤ |z − z0 | 2 2 |dζ| δ γ

61

Since φ is continuous on compact set γ, |φ(ζ)| ≤ M , for a suitable constant M (bounded) Thus |f1 (z) − f1 (z0 )| ≤ |z − z0 |(2M )L(γ) R where L(γ) is the lenth of γ (i.e) L(γ) = γ |dζ|.   δ 2 δ For given  > 0, if we choose η < min 2 , 2M L(γ) , then |z − z0 | < η ⇒ |f1 (z) − f1 (z0 )| < . ∴ f1 (z) is continuous at z0 ∈ γ c . Since z0 is arbitrary in γ c , f1 (z)is continuous in each of the region derermined by γ. Next that in the above context, φ can be replaced by any arbitrary continues function on γ and the corresponding f1 (z) is continuous. We use the continuity of f1 (z) to obtain f10 (z) = f2 (z) Take φ(ζ) =

φ(ζ) ζ−z0

where z0 ∈ γ c . so φ1 (ζ) is still continuous on γ and we have

already proved that Z G1 (z) = γ

φ(ζ)dζ is continuous at z0 (ζ − z0 )(ζ − z)

and so G1 (z) → G1 (z0 ) as z → z0 Now f1 (z) − f1 (z0 ) = G1 (z) z − z0 Thus f10 (z0 ) = lim

z→z0

f 1(z) − f1 (z0 ) = lim G1 (z) = G1 (z0 ) = f2 (z0 ) z→z0 z − z0

Thus completes the proof for n = 1. Let us now assume that fn−1 (z) is analytic and 0 fn−1 (z) = (n − 1)fn (z).

Consider  1 1 fn (z) − fn (z0 ) = φ(ζ) − dζ (ζ − z)n (ζ − z0 )n γ Z



62

Taking 1 z − z0 1 + = n n−1 (ζ − z) (ζ − z) (ζ − z0 ) (ζ − z0 )n (ζ − z0 ) Z Z Z z − z0 φ(ζ)dζ φ(ζ)dζ φ(ζ)dζ + − fn (z) − fn (z0 ) = n−1 (ζ − z ) n n 0 γ (ζ − z) (ζ − z0 ) γ (ζ − z0 ) γ (ζ − z) Then by induction we have Z Gn−1 (z) = γ

φ(ζ)dζ (ζ − z)n−1 (ζ − z0 )

φ(ζ) ζ−z0

which is also continues on γ and G0n−1 (z0 ) = n − 1Gn (z0 ) Z φ(ζ)dζ fn (z) − fn (z0 ) = Gn−1 (z) − Gn−1 (z0 ) + (z − z0 ) (2.16) n γ (ζ − z) (ζ − z0 )

Replacing φ(ζ) by

Using |ζ − z0 | ≥

δ 2

and |ζ − z| ≥ δ for all

Then Z |fn (z) − fn (z0 )| ≤ |Gn−1 (z) − Gn−1 (z0 )| + |z − z0 | γ

≤ |Gn−1 (z) − Gn−1 (z0 )| +

2 δ n+1

|φ(ζ)||dζ| |ζ − z|n |ζ − z0 |

M l(γ)

Both the terms in RHS is small as |z − z0 | is sufficiently small. Note that Gn−1 (z) is analytic at z0 ⇒ Gn−1 (z) is continuous at z0 and

2 M l(γ) δ n+1

is a constant. Thus

fn (z) is continuous. Defining Z Gn (z) = γ

φ(ζ)dζ (ζ − z)n (ζ − z0 )

Then (2.16) can be written as fn (z) − fn (z0 ) Gn−1 (z) − Gn−1 (z0 ) = + Gn (z) z − z0 z − z0 Taking z → z0 in (2.17) we get fn0 (z0 ) = G0n−1 (z0 ) + Gn (z0 ) = (n − 1)Gn (z0 ) + Gn (z0 ) = nGn (z0 ) = nfn+1 (z0 )

63

(2.17)

Note: An analytic function has derivatives of all orders which are analytic and represented by Z

n! f (z) = 2πi n

f (ζ) dζ (ζ − z)n+1

γ

Theorem: 12 (MORERA’S THEOREM). If f (z) is defined and continuous in a R region Ω and if γ f (z)dz = 0 for all closed curve γ in Ω,then f (z) is analytic in Ω. Proof. Let z0 be fixed point and z any variable in Ω and let γ1 , γ2 be any two rectifiable curves in Ω joining z0 to z. Let γ be denote the closed rectifiable curve consisting γ1 and −γ2 so that according to given condition Z Z Z f (z)dz = f (z)dz + f (z)dz = 0 γ γ1 −γ2 Z Z ⇒ f (z)dz = f (z)dz γ1

γ2

This show that integral along every curve in Ω joining z0 to z is the same. Hence, taking r as variable of integration, we may write Z z f (z) = f (ζ)dζ

(2.18)

z0

Notation used in (2.18) for integral is justified in view of the fact that (2.18) is path independent from z0 to z and depents on z0 and set only. Let z + h be any opint in Ω near to z. Then using (2.18) Z

z+h

F (z + h)

f (ζ)dζ

(2.19)

z0

Subtracting (2.19) from (2.18), we get Z

z+h

F (z + h) − f (z) =

Z

z

− z0

Z

z+h

= z0

Z +

z0

z0

Z =

z

z+h

f (ζ)dζ

(2.20)

z

(2.20) is path independent and may be taken along straight line segment joining z

64

to z + h. Hence Z F (z + h) − f (z) 1 z+h = f (ζ)dζ h h z Z  1 z+h 1 F (z + h) − f (z) − f (z) = f (ζ)dζ − hf (z) h h z h Z z+h Z 1 f (z) z+h = f (ζ)dζ − dζ h z h z Z 1 z+h [f (ζ) − f (z)]dζ = h z Since f (ζ) is continues at z, given  > 0 there exist δ > 0, such that |ζ − z|, δ ⇒ |f (ζ) − f (z)| < 

(2.21)

We now choose h 3 |h| < δ, then (2.21) is satisfied for every ζ on the line segment joining z to z + h. Hence, we obtain Z z+h F (z + h) − F (z) 1 |f (ζ) − f (z)||dζ| − f (z) < h |h| z |h| < < |h| Since  is arbitrary we get, F (z + h) − F (z) = f (z) h→0 h lim

⇒ F 0 (z) exist and F 0 (z) = f (z) (primitive) f (z) posses derivative f (z) in every point Ω and consequently F (z) is analytic in Ω. We know that since derivative analytic function is also analytic ⇒ f (z) is analytic. Theorem: 13 (Lioville’s Theorem:). A function which is analytic and bounded in a whole plane must reduce to a constant. Proof. Let z1 , z2 be any two points in z-plane. Let Γ be a circle with center z1 and radius r and z2 is the interior to Γ. Then by cauchy’s integral formula we have Z Z 1 f (ζ)dζ 1 f (ζ)dζ f (z2 ) − f (z1 ) = − 2πi (ζ − z2 ) 2πi (ζ − z1 ) Γ

Γ

65

1 f (z2 ) − f (z1 ) = 2πi

Z

(z2 − z1 )f (ζ)dζ (ζ − z2 )(ζ − z1 )

(2.22)

Γ

Choose R so large that |z2 − z1 | <

R . 2

Then since |z − z1 | = R we have

|z − z2 | = |z − z1 | + |z2 − z1 | ≥ |z − z1 | − |z2 − z1 | ≥ R −

R R = 2 2

Also |f (z)| ≤ M (Bounded). Hence from (2.22 ) we get, Z |z2 − z1 | |f (ζ)||dζ| |f (z2 ) − f (z1 )| ≤ 2π Γ |ζ − z1 ||ζ − z2 | Z |z2 − z1 |2M ≤ |dζ| 2π(R2 ) Γ 1 M × 2 × 2πR × |z2 − z1 | ≤ π R 2M |z2 − z1 | ≤ R Letting R → ∞, We see that RHS → 0 and consequently f (z2 ) = f (z1 ) ∴ f (z) is a constant. Cauchy Estimate: Let f (z) be analytic in a region Ω. Let Ω contain interior and boundary of the circle Γ defined by |z − z0 | = ρ. If f (z) is bounded on Γ, then |f n (z0 )| ≤

n!M . ρn

Proof. We know that Z f (z)dz n! f (z0 ) = 2πi Γ (z − z0 )n+1 Z n! |f (z)||dz| (n) |f (z0 )| ≤ 2πi Γ |z − z0 |n+1 (n)

Since f is bounded, |f (z)| ≤ M. ⇒ |f

(n)

Z n! dz (z0 )| ≤ M n+1 2π Γ |z − z0 | Z M n! 1 M n! 1 dz = × 2πρ ≤ n+1 2π ρ 2π ρn+1 Γ M n! < ρ 66

Theorem: 14 (Fundamental theorem of algebra). Every non-constant polynomial p(z) with complex coefficients has atleast one root. Proof. Suppose p(z) 6= 0 in the whole complex plane C. ⇒ p(z) is analytic and thus

1 P (z)

is analytic in C.

If p(z) = a0 + a1 z + a2 z 2 = · · · + an z n (an 6= 0) By ∆ inequality, a a 0 n−1 + · · · n |p(z)| = |z | an + z z   |an−1 | |a0 | n ≥ |z | |an | − − · · · − n (z 6= 0) |z| |z | n

1 As RHS → ∞ as |z| → ∞, |p(z)| → 0 as |z| → 0 and

|z| ≤ M.

1 P (z)

1 P (z)

is bounded for |z| ≥ M in

being a continuous function is still bounded and so

and analytic in entire complex plane C. By Liouville’s theorem,

1 P (z) 1 p(z)

is bounded = constant.

⇒ p(z) =constant. which is contradicting the fact that p(z) is an n-th degree polynomial. Therefore, P (z) has atleast one zero. Local properties of Analytic functions: Removable singularities: Let us assume f (z) is analytic in 0 < |z − a| < δ for a fixed element a ∈ C. We explicitly assume f is not analytic at z = a. Definition: 21. Let f (z) be analytic in 0 < |z − a| < δ for some δ > 0 and be not analytic at z = a. Then z = a is called an isolated singularity of f. Types of isolated singularities: Let f (z) be analytic in 0, |z − a| < δ for some δ > 0. Let z = a be an isolated singularity of f.

67

• If there exist an analytic function, g(z) in |z − a| < δ such that g(z) = f (z) for z 6= a. We call z = a as a removable singularity of f. • If lim f (z) = ∞, then z = a is a pole for f. z→a

• If z = a is neither a removable or a pole singularity, then it is called isolated essential singularity. Theorem: 15 (Taylor’s theorem). If f (z) is analytic in a region Ω containing a, it is possible to write f (z) = f (a) + a)n−1 + · · · +

fn (z) (z n!

f 0 (a) (z 1!

− a) +

f ” (a) (z 2!

− a)2 + · · · +

f n−1 (a) (z (n−1)!

−

− a)n + (z − a)n fn (z) where fn (z) is analytic in Ω.

Proof. Let z be any point inside the circle C0 with centre a and radius ρ0 . Let |z − a| = ρ0 and C be the circle with center a and radius ρ such that r < ρ < ρ0 , so that z lies inside C. By Cauchy’s integral formula, Z f (ζ)dr 1 f (z) = 2πi C ζ − z and taking    −1 1 1 1 1 1 z−a = = = 1− ζ −z (ζ − a) − (z − a) ζ − a 1 − z−a ζ −a ζ −a ζ−a   2  n−1  n  1 z−a z−a z−a z−a 1 = 1− + + ··· + + . ζ −a ζ −a ζ −a ζ −a ζ −a 1 − z−a ζ−a 1 1 z−z = + + ··· + ζ −z ζ − a (ζ − a)2 Multiplying each term in (2.23) by



f (ζ) 2πi

(z − a)n−1 (ζ − a)n



 +

z−a ζ −a

n .

1 −z + a

(2.23)

and integrating over c and using f (z) be

obtain. 1 2πi

Z c

Z Z f (ζ) 1 f (ζ)dζ (z − a) f (ζ)dζ dζ = + + ··· 2 r−z 2πi c ζ − a 2πi c (ζ − a) Z Z (z − a)n−1 f (ζ)dζ (z − a)n f (ζ)dζ + + n n 2πi 2πi c (ζ − a) c (ζ − z)(ζ − a)

f (z) = f (a) + (z − a)f 0 (a) + (z − a)2

f ” (a) (z − a)n−1 (n−1) + ··· + f (a) + Rn(2.24) 2! (n − 1)! 68

where (z − a)n Rn = 2πi

Z

f (ζ)dζ (ζ − z)(ζ − a)n

c

It is enough to show that Rn → 0 as z → ∞. To prove this we observe that |z−a| = r and |ζ − a| = ρ and therefore |ζ − z| = |ζ − a − (z − a)| ≥ |ζ − a| − |z − a| |ζ − z| ≥ ρ − r Hence when M denote max f (ζ) on C, we get Z |z − a|n |f (ζ)||dζ| |Rn | ≤ n 2π C |ζ − z||ζ − a| rn M 1 ≤ × n × 2πρ 2π ρ − r ρ  n r Mρ r r (since r < ρ ⇒ < 1, → 0 as n → ∞) ≤ × ρ−r ρ ρ ρ ⇒ Rn → 0 as n → ∞. Thus as n → ∞ , the limit is the sum of the first n terms on RHS of (2.24) is f (z). Tthereforef (z) is represented by infnite series f (z) = f (a) +

∞ X (z − a)n n=1

n!

f (n) (a)

known as Taylor’s series. Zeros and poles: If f (a) and all derivatives f (n) vanish, then f (z) = fn (z)(z − a)n for any n, where Z f (ζ)dζ f (z) = f (a) n c (ζ − z)(ζ − a) The disc with circumference C has to be contained in Ω, in which f (z) is defined and analytic . The absolute value |f (z)| has a maximum M on C. If the radius of C is denoted by R we find, |ζ − z| ≥ |ζ − a| − |z − a| ≥ R − |z − a| where |ζ − a| = R 69

M 1 2πR × × n 2π R − |z − a| R M |fn (z)| ≤ n−1 R (R − |z − a|) M ∴ |f (z)| ≤ |z − a|n n−1 R (R − |z − a|)

⇒ |fn (z)| ≤

 |f (z)| ≤

|z − a| R

n

MR R − |z − a|

Hence f (z) → 0 as n → ∞. Theorem: 16. Suppose that f (z) is analytic in region Ω0 obtained by omitting a point 0 a0 from the region Ω. A necessary and sufficient condition that there exist an analytic function in Ω coincides with function in Ω0 that lim f (z) = 0, the extended z→a

function is uniquely determined. Proof. Suppose f (z) has a removable singularity, there exist a analytic function g(z) in |z − a| < δ ⇒ g(z) = f (z) for z 6= a. Then, g(a) = lim g(z) = lim f (z) and z→a

z→a

lim (z − a)f (z) = lim (z − a)g(a) = 0.

z→a

z→a

Conversely, suppose lim (z − a)f (z) = 0.

z→a

Then, a is an exceptional point for f (z). Cauchy’s integral formula is valid indide R f (ζ)dζ 1 where C is the circle with centre at 0 < |z − a| < δ and we have f (z) = 2πi c ζ−z a.We have g(z) =

1 2πi

R

f (ζ)dζ ζ−z

is analytic in the interior of C with its value at z = a become Z f (ζ)dζ 1 g(a) = 2πi c ζ − a R f (ζ)dζ 1 Thus g(z) is analytic in |z−a| < δ and g(z) = f (z) for z 6= a and g(a) = 2πi . C ζ−a C

This proves z = a is a removable singularity for f (z). Furthere f (z) = g(z) for z 6= a implies lim f (z) = lim g(z) = g(a)

z→a

z→a

70

The extended g is unique since g1 and g2 are two extensions, g1 (a) = lim f (z) = z→a

g(a) = g2 (a). Result: We now show that f (z) ≡ 0 in all of Ω. Let E1 be the set in which f (z) and all its derivatives vanish different from 0. E2 is the set where f or one of its derivatives are different from 0. Hence E1 is open and E2 is open becomes f and its derivatives are continuous. Therefore, either E1 or E2 must be empty if E1 is empty, f (z) ≡ 0. If E1 is empty, then f (z) can never vanish with all its derivatives assume f (z) 6= 0. If f (a) = 0, there exist a derivative f (h) (a), which is different from zero. We say that a is zero of order h. An analytic function has the same local behaviour as the polynomial. As in the case of polynomial it is possible to write, f (z) = (z − a)h fn (z), where fh (z) is analytic function and fn (a) 6= 0. Since fn (z) is continuous and fn (a) 6= 0 is a neighbourhood of a and z = a is the only zero in this neighbourhood. The zeros of analytic function which does not vanish identically are isolated. If f (z) and g(z) are analytic in Ω and if f (z) = g(z), on a set which as accumulation point in Ω, then f (z) is identically equal to g(z) (i,e) f (z) = g(z). Theorem: 17. f (z) has a pole of order m. Proof. Consider a function f (z) that is analytic in neighbourhood of a except at a. In otherwords, f (z) is analytic in the region 0 < |z − a| < δ, the point a is an isolated singularity of f (z). If lim f (z) = ∞, point a is called pole of f (z) and set f (a) = ∞. There exist a z→a

0

δ ≤ δ such that f (z) 6= 0 for 0 < |z − a| < δ 0 . In this region, g(z) =

1 f (z)

is defined

and analytic, but the singularity of f (z) at a is removable and g(z) has an analytic extension with g(a) = 0. Since g(z) does not vanish identically to zero at a, has a

71

finite order and we can write, g(z) = (z − a)h gn (z), gn (a) 6= 0 n is the order of pole. f (z) has representation

1 f (z)

1 = (z − a)h h(z)

⇒ f (z) = (z − a)−h fh (z) where fn (z) =

1 gn (z)

is analytic and different from zero in a neighbourhood of a. Meromorphic function: A function f (z) is analytic in a region Ω except for poles is said to be meromorphic in Ω. More perecisely to every a ∈ Ω, there exist a neighbourhood |z − a| < δ contained in Ω suc that either f (z) is in the whole neighbourhood 0 < |z − a| < δ and the isolated singularity is a pole. Note: • The poles of meromorphic function are isolated. • The quotient

f (z) g(z)

of two analytic function in Ω is a meromorphic in Ω provided

that g(z) 6= 0. The only possible poles are the zeros of g(z), but a common zero of f (z) and g(z) can also be removable singularity. • The sum, product and quotient of two meromorphic function are meromorphic. The case of an identically vanishing denominator must be excluded unless we wish to consider ∞ as a meromorphic function. Result: Let f (z) be analytic in 0 < |z − a| < δ. 1. lim |z − a|α |f (z)| = 0 where α is real. z→a

2. lim |z − a|α |f (z)| = ∞ where α is real. z→a

Then, one of the following condition hold: a (i) holds for all real α, f (z) = 0 72

b (i) holds for some α and (ii) for some α and ∃ a unique integer m (i)holds for α > m and (ii) holds for α > m c Neither (i) nor (ii) holds for any real α. Proof. We observe that (i) holds if f (z) ≡ 0 for every α, and conversely if (i) holds for all real α then we claim f (z) ≡ 0, if not by Taking α = 1 lim |z − a||f (z)| = 0

(2.25)

z→a

F (z) has a removable singularity at z = a which has its value is f (a) = lim f (z) = lim (z − a)0 f (z) = 0 (since for any real α (2.25)is true) z→a

z→a

Thus a is zero of f (z) of finite order k, and f (z) = (z − a)k h(z) with h(z) analytic at a and h(a) 6= 0. Thus ⇒ lim (z − a)−kn f (z) = lim h(z) = h(a) 6= 0 z→a

z→a

This is a contradiction since for all real values it should be zero. We note (ii) holds for all α cannot happen because when in this case taking α=0 1 f (z)

is defined in a small neighbourhood of a and (i) holds for

1 . f (z)

Thus f (z) = ∞

cannot happen Next we assume f (z) 6= 0. Let (i) holds lim |z − a|α |f (z)| = 0 for some α, Then ∃ a integer m, z→a

lim |(z − a)|m f (z) = 0

z→a

⇒ lim (z − a)m+1 f (z) = 0 z→a

(z − a)m f (z) has a removable singularity at z = a and its extended value is zero. 73

Thus (z − a)m f (z) has a zero of finite order at z = a. so (z − a)m f (z) = (z − a)k h(z) where h(z) is analytic at a and h(a) 6= 0. Hence lim (z − a)α f (z) = z→a

α+k−m

lim (z − a)

z→a

h(z) = lim (z − a)α−(m−k) h(z) z→a



0 if α > m − k ∞ if α < m − k

= Now let (ii) hold

lim (z − a)α f (z) = ∞ for some α

z→a

Then ∃ an integer m > 0 lim (z − a)m f (z) = ∞

z→a

(z − a)m f (z) has pole at z = a of finite order m, and so (z − a)m f (z) = (z − a)−l k(z) Where k(z) is analytic at a and k(a) 6= 0. Hence lim (z − a)α f (z) = lim (z − a)α−(l+m) k(z)

z→a

z→a

 =

0 if α > m − k ∞ if α < m − k

Wierstrass Theorem for Essential Singularity: An analytic function comes arbitrarily close to any complex value in every neighbourhood of an essential singularity. Proof. Let f (z) be analytic in 0 < |z − a| < δ with z = a as an essential singularity. Let A be any complex no, f (z) comes as close to A as decided in arbitrary neighbourhood of a (i,e) A ∈ C , > 0,δ > 0 ∃z0 with |z0 − a| < δ and |f (z0 ) − A| <  74

If this is not true, then ∃ atleast ane A ∈ C under  > 0 and a δ > 0∃,|z − a| < δ ⇒ |f (z) − A| ≥  Thus lim |z − a|−1 |f (z) − A| = ∞. z→a

So that (ii) of the above result is satisfied for the function f (z) − A with α = 1. Then lim |z − a|α |f (a)| = 0 for some α > 0

z→a

Now, α lim |z − a| |f (z)| = lim (z − a) (f (z) − A + A) z→a z→a α α ≤ lim (z − a) (f (z) − A) + lim (z − a) A α

z→a

z→a

Thus |z − a|α |f (z)| = 0 for some α and so z = 0 is not an essential singularity. Which is a contradiction.   1 Sum Describe the Remann surface associated with function ω = z+z   = 21 1 − z12 which vanishes when z ± 1. We observe ω → ∞ as z → 0, and dω dz 1 2

Hence z = −1 and z = 1 are critical points of the transformation. now, ω = U + iV and z = r(cosθ + isinθ) the transformation can be written as   1 1 u + iv = r(cosθ + isinθ) + (cosθ − isinθ) 2 r   1 1 1 = (r + )cosθ + i(r − sinθ) 2 r r Hence equating real and imaginary parts we get 1 1 1 1 u = (r + )cosθ and v = (r − sinθ) 2 r 2 r

(2.26)

(i)We consider the image of circle |z| = λ, (i, e)r = λ. Sub this and eliminating θ from (2.26), the image curve is given by v2 u2 + 1 1 2 1 1 2 = 1 (λ + ) (λ − ) 4 λ 4 λ 75

(2.27)

Which represent family of ellipses for varying values of λ with semi-axis a = 12 (λ+ λ1 ) and b = 12 (λ + λ1 ) The eccentricity of these ellipses is given by 1 1 1 1 (λ − )2 = (λ − )2 (1 − e2 ) 4 λ 4 λ This gives e = 1 ) λ

+

2 1 ) (λ+ λ

2 1 ) (λ+ λ

hence the foci of the points ω = ±ae (i,e) ω = ±ae = ± 12 (λ =

which are independent of λ. Thus (2.27) represents a family of confocal

ellipses with foci at the points ω = ±i. We examine these ellipses more closely. When r = 1, ((2.26)) gives u = cosθ and v = 0, Hence as z goes round the circle |z| = 1 and θ varies from 0 and 2π and u moves from point w through 0 point −1 and then comes back to the point 1 itself. In other words, the ellipse corresponding to unit circle degenerates to points ω = −1 to ω = 1 We now consider the image of |z| = λ1 = r where λ1 > 1 then the image of ellipse, U2 1 (λ1 + 4

1 ) λ1

+

V2 1 (λ1 − 4

1 ) λ1

=1

It is evident that (2.28) is also image of circle |z| = r =

(2.28) 1 λ1

as λ1 increases from 1 to

∞ and decreases 1 to 0. λ1 +

1 λ1

and λ1 +

1 λ1

constantly increases. It follows that each of the 2 ring

shaped regions λ1 < |z| < λ2 , λ12 < |z| <

1 λ1

(λ2 > λ1 ) in the z-plane is mapped

conformally on the region bounded by the 2 confocal ellipses represented by (2.28) and λ1 replaced by λ2 in (2.28).

Again making λ1 → ∞ we observe that each of regions |z| > λ1 and |z| < 76

1 λ1

is

conformally mapped onto the region exterior to ellipse (2.28). Now making λ1 → 1 we see that region |z| > 1 and |z| < 1 are mapped conformally on whole w-plane with a slit along real axis −1 to 1 as

(ii)We next consider the image of radial line θ = α for varying α, subs θ = α and eliminating r from (2.26) we obtain, u2 v2 − =1 cos2 α sin2 α

(2.29)

Which represent a family of convocal hyperbolas with vertices at the point ω = ±cosα and the foci at ω = ±1. Thus the foci of these hyperbolas coincide with those of ellipses in (2.26). Since 1 1 u = (γ + )cosα; 2 γ

1 1 v = (γ − )sinα 2 γ

We see that as r increases from 1 to ∞ on the ray 0 = α. The point (u, v) starting from the vertex (cosα, 0) moves on the upper half of the hyperbola (2.29) lying on the RHS of imaginary axes, Again as r decreases from 1 to 0 on ray θ = α The point (u, v) starting from (cosα, 0) move on lower half of right hand branch of (2.29) Thus the ray θ = α corresponds to the complex right hand branch of (2.29). The same branch is also complete image of the ray θ = 2π − α which we can verify. Similarly it can be shown that the left hand branch of hyperbola is the complete image of either of 2 rays θ = π − α, θ = π + α. It follows that the edge shaped regions α < θ < β and 2π − β < θ < 2π − α are both mapped conformally on the 77

domain bonded by the 2 branches lying on the right of image axes of hyperbola U2 V2 − = 1, cos2 α sin2 α

U2 V2 − =1 cos2 β sin2 β

It is evident that as θ is corresponding branch of hyperbola moves towards the origin when θ =

π 2

we have u = 0, v = 21 (r − 1r ). As varies form 0 to ∞ on θ = π2 ,

It is under that point (uv) describes complete image axis in w-planes. θ =

3π 2

also

corresponds to whole of image axis in w-plane. Thus each of region defined by α<θ<

3π π and < θ < 2π − α 2 2

is mapped conformally on the region bounded by image axes and the branch of hyperbola (2.29) on the right. Similarly each of the regions defined by and π + α < θ <

3π 2

π 2

< θ < π −α

is mapped conformally on the region bounded by image axes

and branch of (2.29) on the left. If follows that each of the regions, defined by π + α < θ < 2π − α & α < θ < π − α is mapped conformally on the region in w-plane bounded by 2 branches of hyperbola (2.29). We now examine the interior of hyperbola. When θ = 0. We have u = 21 (r + 1r ), v = 0. As r varies from 0 to ∞, it is easy to verify that point(u, v) describes the segment of +ve real axis from 1 to ∞ twice. Hence as θ decreases from α to 0, we see that the wedge 0 < θ < α is mapped on the interior of right hand branch of hyperbola (2.29) with the slit along the +ve real axis from 1 to ∞. Observe that the wedge shaped regions defined by : 2π − α < θ < 2π also corresponds to the same interior of the hyperbola as shown 78

below: similarly each of region defined by π − α < θ < π & π < θ < π − α is

mapped conformally to the interior of left hand branch of (2.29) with a slit along the -ve real axis from −1 to ∞. We this conclude that the whole z-plane is mapped conformally into 2 − 1 correspondence onto. The whole w-plane with 2 slits on real axis extending from 1 to ∞ and −1 to −∞ respectively (iii)Now each ray θ = α cuts every circle r = δ orthogonally since the mapping is conformal, the 2 families of ellipses and hyperbola in (i) and (ii) must also form a orthogonal system which is also a well-known of analytical geometry. These families are shown in the fig:

Theorem: 18. Let zj be zeros of the function f (z) which is analytic in a disc ∆ and does not vanish identically. Each zero being counted as many times as its order indicates. For every closed curve γ in ∆ which does not pass through a zero Z 0 X f (z) 1 nj (γ, zj ) = dz 2πi γ f (z) j where the sum has only a finite no. of terms 6= 0 Proof. We observe that the zeros of f (z) in ∆ cannot have limit points in ∆. Thus in the region enclosed by γ, there can have finite zeros of f (z) by contrapostive argument of Bolzano-Wierstrass theorem. Then n(γ, zj ) 6= 0

only for finitely many zj

79

Now γ is contained in disc C concentric with ∆ such that c¯ contained in ∆. (Let ∆ be disc with centre a, radius r and observe d = sup{|z − a|} < r and choose one z∈γ

ρ there exists d < ρ < r. Then |z − a| < ρ is the required ρ. By repeating the argument with c¯ in the place of γ, we again see that c¯ and interior of c contains only finitely many zeros of f (z). Let these finite zeros be written as {z1 , z2 , · · · , zn } where we repeat each zj as many times as order indicates. Then, f (z) = (z − z1 )(z − z2 ) · · · (z − zn )[g(z)] where g(z) is analytic in interior of c and has no zeros. Thus log f (z) = log(z − z1 ) + log(z − z2 ) + · · · + log(z − zn ) + log g(z) 1 1 1 g 0 (z) f 0 (z) = + ··· + + f (z) z − z1 z − z2 z − zn g(z) Z 0 Z Z Z 0 1 1 1 1 f (z) 1 1 g (z) dz = dz + · · · + dz + dz 2πi f (z) 2πi z − z1 2πi z − zn 2πi g(z) γ γ γ γ Z 0 Z 0 f (z) 1 g (z) 1 dz = n(γ, z1 ) + · · · + n(γ, zn ) + dz 2πi f (z) 2πi g(z) γ γ X = n(γ, zj ) j

where

1 2πi

R

g 0 (z) dz γ g(z)

= 0. Since

g 0 (z) g(z)

is analytic in ∆.

Theorem: 19 (Local correspondence theorem). Suppose that f (z) is analytic at z0 and f (z0 ) = ω0 and that f (z) − (ω0 ) has zero of order n at z0 . If  > 0 is sufficiently small there exist a corresponding δ > 0, ∀a with |a − z0 | < δ the equation f (z) = a has exactly n roots in the disc |z − z0 | < . Proof. Let f (z) − ω0 have a zero of order n at z0 . Choose  > 0 so that f (z) is defined and analytic in |z − z0 | <  and also such that z0 is the only zero of f (z) − ω0 is that dosed disc |z − z0 | < . Let γ be the simple closed curve |z − z0 | =  and Γ is its image under f . now ω0 ∈ Γ and so ω0 ∈ Γc . Thus there exist one and only region ξ determined by Γ containing Γ and containing ω0 . using the properties of 80

the index n(Γ, a) it is clear that n(Γ, ω0 ) = n(Γ, ω) for every ω ∈ ξ but 1 n(Γ, ω0 ) = 2πi 1 = 2πi

Z ZΓ

dω ω − ω0 f 0 (z)dω since ω = f (z) f (z) − ω0

γ

By applaying previous theorem, for f (z) − ω0 for f (z), n(Γ, ω0 ) =

X

n(γ, aj )

j

where aj ’s are the zeros of f (z) − ω0 inside Γ. But the no.of zeros of f (z) − ω0 inside Γ is precisely n zeros. Since n(γ, aj ) = 1 or 0 depending on which aj is inside or P outside of γ. We have n(γ, aj ) = n. Thus j

n(Γ, ω0 ) = n further n(Γ, ω) =

P

n(γ, bj ) where we ξ and bj are zero of f (z) − ω.

j

(i,e) No.of zeros of f (z) inside Γ is equal to n. on the Other hand, If  > 0 it is chosen a |z − z0 | ≤  is free of zero of f (z) also. The solutions of f (z) − ω in |z − z0 | <  are also simple. However |z − z0 | <  is connected and so its image under cont. map f must lie in ξ only. Thus between |z − z0 | <  and the corresponding ξ the mapping f is n − 1 (n to one). Corollary 1 (Open Mapping Principle). A non constant analytic function maps open set onto open sets. Proof. Let f (z) be a non-constant analytic function in Ω. Let U be an open subset of Ω and z0 ∈ U with f (z0 ) = ω0 there exist a +ve integer n ≥ 1, for which f (z) − ω0 81

has zeros of order n. By local correspondence theorem, ∃ > 0 and ξ containing ω0 , |z − z0 | <  is a subset of U and each ω ∈ ξ is assumed by f at n points in |z − z0 | < . Thus ∃ atleast one point z ∈ U with f (z) = ω. Since ξ is open and ω0 ∈ ξ , ∃ a δ > 0, |ω − ω0 | < δ is a subset of ξ. Hence |ω − ω0 | < δ is completely contained in f (U ). (i.e) ω0 is an interior point of f (U ). Since ω0 is arbitrary, ∴ f (U ) is open. Corollary 2. If f (z) is analytic at z0 with f 0 (z0 ) 6= 0, it maps a neighbourhood of z0 conformally and topologically onto a region. Proof. By local correspondence theorem, ∃ a  > 0 and corresponding region ξ containing ω0 such that |z − z0 | <  maps one-one and onto to ξ. Since f 0 (z0 ) 6= 0 ⇒ z0 is a simple zero of f (z) − ω0 . This map is obviously continuous and by open mapping principle, this is also open. Thus both f and f −1 are continuous and so the correspondence is topological. Since f 0 (z) 6= 0 in |z − z0 | < , the correspondence is conformal. Theorem: 20 (Maximum principle). If f (z) is analytic and non-constant in a region Ω, then its absolute value |f (z)| has no maximum in Ω Proof. Let z0 ∈ Ω and ω0 = f (z0 ) be its image. By open mapping principle, there exist  > 0 and δ > 0 such that every point ω with |ω − ω0 | <  is the image of some point z in |z − z0 | <  (i,e) f (z) = ω. Since in this neighborhood |ω − ω0 | < δ of ω0 we can always find points. ω with |ω| > |ω0 |, |ω0 | cannot be the maximum of all |f (z)| as z-varies in Ω, since z0 is arbitrary, there cannot be any z0 ∈ Ω with |ω0 | = |f (z0 )| = max |f (z)|. z⊂Ω

Theorem: 21 (MAXIMUM MODULUS THEOREM). If f(z)is defined and continuous on a closed and bounded E and analytic on interior of E, then the maximum of |f (z)| is attained on its boundary of E. Proof. Let E be a compact set and the derived set E 0 6= φ and f(z) is analytic on E 0 and continuous on E. Since E is compact and |f (z)| is continuous, ∃ a point z0 ∈ E 82

3 |f (z0 )| = max |f (z)| z⊂Ω

If z0 is on the boundary of E, then there is nothing to prove. Let z0 ∈ E 0 , Let A be one of the components of E 0 3 z0 ∈ A since A is open (components of open sets are open and E 0 is open), we can find r > 0 3 N = {z : z|z − z0 | < r} ⊆ A. A is open, connected and non-empty. ⇒ A is region. ∴ |f (z0 )| is also the max.|f (z)| in A and as such it attains its maximum at one of the interior points, namely z0 . By Maximum principle, we see that f(z) reduces to a constant (i.e.,) f (zo ) in A and by continuity f(z) is also the same constant on the boundary of A. Let a be point on boundary of A...(*) Let E be the compact set and A be one of the components (open) of interior of E (E 0 ), then each boundary point of A is also a boundary point of E. ⇒ a is boundary point of E also. Note that ∂A 6= φ = A¯ ∩ A¯c . Then f (z) = f (z0 ) for every z ∈ A¯ and |f (a)| = |f (z0 )| = max |f (z)| z∈E

since (*) ⇒ a is boundary point of E. If E 0 is connected A = E 0 and if z0 is an interior point of E, then f (z) ≡ f (z0 ) in E 0 . On the other hand if f is analytic in a region Ω, E ⊂ Ω, z0 is interior to E then over a subregion A of Ω, f (z) ≡ f (z0 ) and by principle of analytic continuation, f (z) ≡ f (z0 ) in Ω. since f (z0 ) is constant, it is the maximum of |f (z)|. SCHWARZ LEMMA: If f(z) is analytic for |z| < 1 and satisfies the condition |f (z)| ≤ 1, f(0)=0 then |f (z)| ≤ |z| and |f 0 (0)| ≤ 1. If |f (z)| = |z| for some z 6= 0 or if |f 0 (0)| = 1, then f (z) = cz for somewith a constant c of absolute value 1.

83

Proof. Define f1 (z) =

f (z) z

for z 6= 0

f1 (0) = lim

z→0

f (z) f (z) − f (0) = lim z→0 z z−0

f1 (0) = f 0 (0) ⇒ f1 is analytic for z 6= 0. Using finite series development at origin, we get f (z) = f (0) + f 0 (0)z + z 2 h(z) ⇒ h(z) =

f (z) − f (0) − zf 0 (0) z2

where h(z) is analytic in |z| < 1. Since f (0) = 0,we have f (z) z

− f1 (0) f (z) − zf1 (0) = lim z→0 z z2 0 f (z) − zf (0) ≤ lim = lim h(z) = h(0) z→0 z→0 z2

f1 (z) − f1 (0) = lim lim z→0 z→0 z−0

∴ f1 is analytic at zero.Now let us take the circle |z| = r < 1 and we observe that f (z) |f (z)| 1 = |f1 (z)| = ≤ z |z| r on |z| = r and hence by maximum modulus theorem, |f1 (z)| ≤

1 r

on |z| = r. Letting

r → 1,we see that |f1 (z)| ≤ 1 for all z in |z| < 1. On the other hand,if equality in |f1 (z)| ≤ 1 is held at any single point in |z| < 1. ∴ By Liouville’s theorem, f1 (z) must reduce to a constant(say c) Clearly |c| = 1 and f (z) = cz. Generalised Schwartz Lemma: Let f (z) be analytic in |z| < R and satisfy |f (z)| < M for |z| < R. Let f (z0 ) = ω0 with |z0 | < R and |ω0 | < M ,then M f (z) − ω0 R(z − z0 ) M 2 − ω¯0 f (z) ≤ R2 − z¯0 z 84

Proof. Let r = |T (z)| be a bilinear transformation which maps |z| < R onto |r| < 1 with T (z0 ) = 0. Let ξ = S(ω) be a bilinear transformation which maps |ω| < M onto |ξ| < 1 with S(ω0 ) = 0. Clearly T −1 maps |r| < 1 to |ξ| < 1 with 0 going to 0 under this map. We also have the explicit transformation r = T (z) = and ξ = S(ω) =

R(z − z0 ) R2 − z¯0 z

M (f (z)−ω0 ) M 2 −ω¯0 f (z)

Using Schwartz lemma,we get |R(z − z0 )| |S ◦ f ◦ T −1 (r)| ≤ |r| = |R2 − z¯0 z| R(z − z0 ) ⇒ |ξ| ≤ 2 R − z¯0 z M (f (z) − ω0 ) R(z − z0 ) ≤ M 2 − ω¯0 f (z) R2 − z¯0 z

CHAINS AND CIRCLES: The equation Z

Z f dz =

γ1 +...+γn

Z f dz + ... +

γ1

f dz

(2.30)

γn

which is valid when γ1 + γ2 + ... + γn form a subdivision of arc γ.Since RHS of (2.30) has meaning for any finite collection,nothing prevents us from considering an arbitrary formal sum γ1 + γ2 + ... + γn which need not be an arc and we define the corresponding integral by (2.30). Such formal sum of arcs are known as chains. Definition: 22 (Cycles). A chain is a cycle if it can be represented as a sum of closed curves. A chain is a cycle iff in any representation the initial and end points of individual arcs are identical in pairs. 85

Result: The integral of exact differential over any cycle is zero. Definition: 23. A cycle γ is an open set Ω is said to be homologous to zero with respect to Ω if n(γ, a) = 0 for all a ∈ Ωc .

86

Chapter 3

CALCULUS OF RESIDUES Theorem: 22 (RESIDUE THEOREM:). cauchy’s integral formula can be expressed as: If f (z) is analytic in a region Ω ,then n(γ, a) =

1 2πi

R γ

f (z)dz z−a

for every cycle γ which

is homologous to zero. The residue of f (z) of at an isolated singularity a, is the R , the derivative of single valued unique complex number R which makes Sf (z) − (z−a)

analytic function in the annulus 0 < |z − a| < δ Theorem: 23 (CAUCHY RESIDUE THEOREM). Let f (z) be analytic except for isolated singularities aj in a region Ω,then Z X 1 f (z)dz = n(γ, aj )Resz→aj f (z) 2πi γ j where n(γ, aj ) is the order for any cycle γ which is homologous to zero in Ω and does not pass through any of aj Proof. Let γ be a cycle which is homologous to zero and does not pass through any of aj . Let cj be a circle with centre aj . R 1 Then the 2πi f (z)dz is defined as residue of f (z) at singularities aj . cj Also Rj is defined by 1 Rj = 2πi

Z f (z)dz cj

87

consider the cycle Γ = γ − n(Γ, ak ) = n(γ −

X

P

j

n(γ, aj )cj . now,

n(γ, aj )cj , ak )

j

= n[γ − (n(γ, a1 )c1 + ... + n(γ, ak )ck ), ak ] = n(γ, ak ) − n(γ, a1 )n(c1 , ak ) − n(γ, a2 )n(c2 , ak )... − n(γ, ak )n(ck , ak ) = n(γ, ak ) − 0 − 0... − n(γ, ak )

(since ak ∈ / cj where j 6= k)

n = [Γ, ak ] = 0 now we take a ∈ / Ω, then n(Γ, a) = n(γ −

X

n(γ, aj )ck , a)

j

X = n(γ) − n( n(γ, aj )ck , a) j

= 0 Therefore Γ is cycle in Ω which is homologous to zero in Ω and does not pass through any of point aj . By general form of Cauchy theorem, Z Z Z f (z)dz = 0 ⇒ f (z)dz = P Γ

γ

Z ⇒

f (z)dz = γ

X X

f (z)dz =

X

f (z)dz n(γ,aj )cj

Z f (z)dz

n(γ, aj ) cj

j

=

j

n(γ, aj )(2πi)Resz=aj f (z)

j

1 2πi

Z γ

n(γ, aj )Resz=aj f (z)

j

suppose there are infinite number of singularities then the above result is true for if the infinite sum in RHS consists only finite number of forms which is non-zero. Also the set of all point aj in which n(γ, aj ) = 0, hence there are only finite number of for which n(Γ, aj ) 6= 0. Definition: 24. A cycle γ is said to bounded line region Ω if and only if n(γ, a) is defined and n(γ, a) = 1 for all a ∈ Ω and either undefined or zero for all a ∈ / Ω. 88

Definition: 25. If γ bounded Ω and Ω + γ is contained in a larger region Ω0 , then it is clear that γ is homologous to zero in Ω0 . Results: If f (z) is analytic in Ω + γ then

R

f (z)dz = 0 and f (z) = γ

1 2πi

R γ

f (r)dz r−z

for all z

in Ω. Therefore if f (z) is analytic in Ω + γ except for isolated singularities in Ω, R P P 1 f (z)dz = j Resz−aj f (z) where ranging over all aj in Ω 2πi γ

Theorem: 24 (ARGUMENT PRINCIPLE:). If f(z) is meromorphic in Ω with zeros aj and pole bk , then

1 2πi

Z γ

X X f 0 (z) dz = n(γ, aj ) − n(γ, bk ) f (z) j k

(3.1)

Proof. In general, there can be infinitely many zeros and poles of f (z) in Ω.It is noted that the sums on RHS of (3.1) are always finite because inside Ω there can atmost finitely many aj ’s and bk ’s as other wise there will be either a limit point of zeros or limit point of poles inside or on γ of which are excluded by our assumption. If aj is a zero of order h, for f (z) then f (z) = (z − aj )b fi (z), in a neighbourhood of aj where f1 (z) is analytic and f1 (aj ) 6= 0. If follows from continuity of f1 (z) at aj . that there is a neighbourhood of aj in which f1 (z) 6= 0. Thus in the neighbourhood. logf (z) = hlog(z − aj ) + logf1 (z) 1 h 1 0 f 0 (z) = + f (z) f (z) z − aj fi (z) Taking Taylor series expansion for g(z) at z = aj g(z) = g(aj ) + (z − aj )g 0 (aj ) + (z − aj )2 Truncating. g(z) = g(aj ) + (z − aj )g 0 (aj ) 89

g 00 (aj ) + ... 2!

where g(z) is an analytic function in Ω. Then, g(z)

f 0 (z) h f 0 (z) b f 0 (z) = g(aj ) + g(aj ) 1 + g 0 (aj )(z − aj ) + g 0 (aj )(z − aj ) 1 + ... f (z) z − aj f1 (z) z − aj f1 (z)

The coefficient of

1 z−aj

0

0

(z) (z) for g(z) ff (z) is hg(aj ) , the residue of g(z) ff (z) at z = aj .

0

(z) Thus g(z) ff (z) has a pole atz = aj with residue hg(aj )

similarly f (z) = of f (z) of order

1 f (z) where f2 is analytic (z−bk )k 2 0 (z) also has a pole with k, g(z) ff (z)

and f2 (bk ) 6= ∞ then at pole bk residue −kg(bk ), Then by using

cauchy residue theorem, we get, Z X X 1 f 0 (z) g(z) dz = n(γ, aj )g(aj ) − n(γ, bk )g(bk ) 2πi f (z) j k take g(z) ≡ 1, then we get Z X X f 0 (z) 1 n(γ, bk ) g(z) dz = n(γ, aj ) − 2πi f (z) j k

Definition: 26 (GENERALISED ARGUMENT PRINCIPLE). If f (z) 6= 0, we define in a region Ω with zeros aj and poles bk with no other singularities in Ω. Let g(z) be analytic in Ω. Let γ be homologous to zero in Ω not passing through a0j s and b0k s.Then 1 2πi

Z g(z) γ

X X f 0 (z) n(γ, bk )g(bk ) dz = n(γ, aj )g(aj ) − f (z) j k

Theorem: 25 (ROUCHE’S THEOREM). Let γ be homologous to zero in Ω and such that n(γ, z) is either 0 or 1 for any point z ∈ / Ω. Suppose that f (z) and g(z) are analytic in Ω and satisfy. |f (z) − g(z)| < |g(z)| on γ, then f (z) on g(z) have same number of zeros enclosed by γ. Proof. In order to prove the above we first, ”Let γ be homologous to zero in Ω such that n(γ, a) = 0 or 1 for a ∈ / γ, Let f (z) and g(z) be analytic in Ω with |g(z)| < |f (z)| on γ(∗) then f (z) and g(z) + f (z) have same number of zeros enclosed by γ”. 90

*proof for above statement: Note that if h(z) = A(z)B(z) where A and B are analytic. ⇒ log h(z) = log A(z) + log B(z) h0 (z) A0 (z) B 0 (z) ⇒ = + h(z) A(z) B(z) we now write f + g = f (1 + g/f ) on γ,then log(f + g) = log f + log(1 + g/f ) (f + g)0 f 0 (1 + g/f )0 = + on f +g f (1 + g/f )

γ

Note that |g(z)| < |f (z)| on γ implies that f has no zeros. on γ. By argument principle we have, Z 1 (f + g)0 (z) dz = Number of f + g enclosed byγ. 2πi γ (f + g)(z) Z 0 1 f (z) dz = Number of zeros of f enclosed byγ. 2πi γ f (z) Now put Γ = (1 + g/f )(γ) and we know that

|g(z)| |f (z)|

< 1. If and only if |1 +

g(z) f (z)

− 1|

on γ, (i.e) Γ lie inside the open disc D of centre 1 and radius 1 and does not around zero. Thus n(Γ, 0) =

1 2πi

dω Γ ω

R

= 0. Since

1 ω

is analytic in D and apply cauchy’s

theorem for a circular disc, then 1 0 = n(Γ, 0) = 2πi

Z γ

dω 1 = ω 2πi

Z γ

(1 + g/f )0 (z) dz (1 + g/f )0 (z)

Hence number of zeros of f + g enclosed by γ Replace g by f − g and f by g, in the second statement in (*) |(f − g)(z)| < |g(z)| ⇒ |f (z) − g(z)| < |g(z)| Therefore by applying the theorem ⇒ we have, number of zeros of f equal to number of zeros of g on γ. 91

9. How many roots does the eqn z 7 − 2z 5 + 6z 3 − z + 1 = 0

(3.2)

have in the disc |z| < 1. Proof. Let f (z) = 6z 3 and g(z) = z 7 − 2z 5 − z + 1 Then |f (z)| = 6|z|3 = 6

(since |z| < 1)

and |g(z)| ≤ |z|7 − 2|z|5 − |z| + 1 < 5. Here |g(z)| < |f (z)| hence by Rouche’s Theorem. f +g and f have the same number of zeros inside |z| < 1. However f (z) has exactly 3 zeros in |z| < 1. therefore (3.2) has exactly 3 zeros in |z| < 1.

10. How many roots of the equation z 4 − 6z + 3 = 0 having their madulus between 1 and 2. Proof. Let f (z) = −6z and g9z) = z 4 + 3 on |z| ≤ 1 |g(z) ≤ |z|4 + 3 ≤ 4

and

|f (z)| ≤ 6|z| = 6

Therefore by Rouches theorem f + g = z 4 − 6z + 3 and f = −6z have same number of zeros in |z| ≤ 1. Number of f (z) =Number of zeros of f + g = 1. in |z| ≤ 1 Now let h(z) = z 4 and k(z) − 6z + 3 on |z| Here, |h(z)| = |z|4 = 16 |k(z)| ≤ 6|z| + 3 ≤ 15 ∴ |k(z)| < |h(z)| Therefore by Rouche’s theorem h+k and h have same number of zeros in |z| < 2 and has just 1 root in |z| < 1. on|z| = 1 → |z 4 − 6z + 3| ≥ 6 − 1 − 3 = 2 > 0. Therefore 11 has no zeros in |z| = 1. Number of roots of z 4 − 6z + 3 lying on 1 < |z| < 2 is precisely 4 − 1 = 3. 92

EVALUATION OF DEFINITE INTEGRAL Integration round the unit circle We consider integral of the type

R2π

f (cos θ, sin θ)dθ. where the integrand is a

0

rational function of sin θ cos θ.If we write 1 1 1 1 z = eiθ , dz = ieiθ dθ, cos θ = (z + ), sin θ( )(z − ) 2 z 2i z Then when γ unit cricle, Z2Π 0

Z 1 1 1 1 dz 1 f (cos θ, sin θ)dθ = lim f ( (z + ), (z − )) iθ γ i 2 z 2i z e Z = Φ(z)dz γ

Example:

11. Evaluate

Rπ 0

dθ a+cosθ

Here, Zπ

1 dθ = a + cosθ 2

0

Z2π

dθ a + cosθ

0

1 1 dz substitute z = eiθ , dθ = iθ , cos θ = (z + ) ie 2 z Z2π Z 1 dθ = a + cosθ 2i 0 γ Z 1 = 2i γ Z 1 = i

dz/eiθ a + 12 (z + z1 ) dz 2 . z (2az + z 2 + 1) z2

γ

Then finding roots of z 2 + 2az + 1,

√

√ a2 − 1 2 √ √ α = −a + a2 − 1; β = −a − a2 − 1 z =

−2a ±

4a2 − 4

93

= −a ±

dz + 2az + 1

since |a| > 1, we get |β| > 1

√ √ αβ = a2 − a a2 − 1 − a a2 − 1 − a2 + 1 = 1

therefore |αβ| = 1 ⇒ |β| = here f (z) =

1 . z 2 +2az+1

1 |α|

> 1 ⇒ |α| < 1

Now we have find the residue at α

α, is a simple pole. 1 1 = lim (z − α) z→α f (z) z→α (z − α)(z − β) 1 1 = √ = α−β 2 a2 − 1 π Z Z Π dθ dz 1 =√ = −i lim 2 = −i(2πi) √ 2 2 γ z + 2az + 1 a + cosθ 2 a −1 a −1

Resz=α f (z) = lim (z − α)

0

a)

π/2 R 0

dθ , |a| a+sin2 θ

>1

Let Zπ/2 I =

dθ ⇒ 4I = a + sin2 θ

Z0 ⇒ 4I =

2π

a+

dθ a + sin2 θ

Z0

dθ

0

Z2π

2θ ( 1−cos ) 2

2π

=

2dθ 2a + 1 − cos 2θ

0

Let 2θ = Φ ⇒ 3dθ = dφ, 0 ≤ φ ≤ π

Z2π 4I =

dφ ⇒ 2I = 2a + 1 − cosφ

Z2π

0

0

Z2π

Z

dφ 2a + 1 − cosφ

dφ 1 dz/z 1 1 = i 4az + 2z − z 2 − 1/z 2a + 1 − 2 (z − z ) 0 γ Z Z 2dz dz 2I = +i = i lim 2 2 γ z − z(4a + 2) + 1 z − (4a + 2)z + 1 2I =

γ

94

here f (z) =

1 z 2 −z(4a+2)+1

finding roots forf (z), √

18a2 + 4 + 16a − 4 √ 2 4a + 2 ± 4 a2 + a = 2√ z = 2a + 1 ± 2 a2 + a

z =

4z + 2 ±

√ √ α = 2a + 1 + 2 a2 + a; β = 2a + 1 − 2 a2 + a |β| < 1 ⇒ β lies inside γ therefore Resf (z) = lim (z − β) z=β

z→β

1 1 = (z − α)(z − β) β−α

1 √ −4 a2 + a Zπ/2 Z 1 π dθ dz p √ = i = i(2πi) = 2 a + sin2 θ 2 a2 + a −4( a2 + a) γ z − (4a + z)z + 1 =

0

R∞

Evalution of integral of the form

f (z)dz.

−∞

Theorem: 26. If f (z) is a function which is analytic in the upper half of the plane, except at a finite number of poles in it,having again no poles on the real axis and if R P further zf (z) → 0 as |z| → ∞, then by contour integration, ∞f (z)dz = 2πi R+ −∞

where R+ : sum of residues of poles in upper half plane. Exercise R∞ x2 dx R∞ 2 b) (x2 +a2 )3 , a is real Let I = (x2x+adx2 )3 since the integral is even. 0

⇒I=

1 2

R∞ −∞

0

P (x) dx (Q(x))

where P (x) = x2 ; Q(x) = (x2 + a2 )3

since deg(denominator)>deg(numerator), we cannot have poles on R here f (z) =

P (z) z2 = 2 Q(z) (z + a2 )3

To find poles of f(z), it is equal of zeros of Q(z)

95

therefore (z 2 + a2 )3 = 0 z 2 + a2 = 0 z = ±ia only α = ia lies in the upper half plane 1 P (z) P (z) (2πi)Resz=ia = πiResz = ia 2 Q(z) [(z − ia)(z + ia)]3 g(z) = πiResz=ia (z − ia)3 00 g (ia) I = πi 2!

⇒I =

consider,g(z) =

z2 (z+ia)3

0

g (z) = ⇒ g 0 (z) = g 00 (z) = = = = g 00 (ia) = I =

(z + ia)2 z[(z + ia)2 − 3z] (z − ia)3 2z − z 2 (3)(z + ia)2 = (z + ia)6 (z + ia)6 (2aiz − z 2 ) z(2ai − z) = z + ia)4 (z + ia)4 (z + ia)4 (2ai − 2z) − 4(2aiz − z 2 )(z + ia)3 (z + ia)8 (z + ia)3 [(z + ia)(2ai − 2z) − 4(2aiz − z 2 )] (z + ia)8 1 [2aiz − 2z 2 − 2a2 − 2iaz − 8aiz + 4z 2 ] (z + ia)5 2(z 2 − a2 − 4aiz) 1 2 2 [2z − 2a − 8aiz] = (z + ia)5 (z + ia)5 2[−a2 − a2 + 4a2 ] 4a2 1 = = 5 5 32ia i32a 8ia3 1 1 π πi( 3 )( ) = 8ia 2! 16a3

along the real line .Now considering the hemisphere γR, then the integral along is Z Z |z|2 lim |f (z)||dz| = lim lim 2 lim |dz| γR R→∞ γR |z + a2 |3 R→∞ Z R2 lim 2 ≤ lim |dz| R→∞ γR (R + a2 )3 R2 πR = lim [ 6 ] R→∞ (R (1 + a2 /R2 )3 π = =0 (1 + a2 /∞) 96

Then the integral, Z∞ =

π 16a3

+∞

R∞ 0

x2 dx x4 +5x2 +6

Let I =

R∞ 0

x2 dx x4 +5x2 +6

Since the function is even, I =

1 2

R∞ −∞

x2 dx x4 +5x2 +6

=

1 2

R∞ P (x) 0

Q(x)

dx, where P (x) = x2 ;

Q(x) = x4 + 5x2 + 6 Here since deg(P (x)) < deg(Q(x)), there can be no poles on R f (z) =

z2 z 4 + 5z 2 + 6

The poles of f (z)=zeros of Q(z)

z 4 + 5z 2 + 6 = 0 √ −5 ± 25 − 24 z2 = 2 −5 ± 1 z2 = 2 2 z = −2, −3 √ √ z = ± 2i, ± 3i

97

√ √ z = i 2, i 3 lies in upper half plane. Resz=i√2

P (z) = Q(z)

√ lim√ (z − i 2)

z→i 2

z2 √ √ (z − i 2)(z + i 2)(z 2 + 3)

−2 √ 2i 2 −1 = √ i 2 =

Resz=i√3

P (z) = Q(z) = = I = = =

√ lim√ (z − i 3)

z→i 3

z2 √ √ (z − i 3)(z + i 3)(z 2 + 2)

−3 √ 2i 3 √ 3 2i 2πi(Residue) √ ! 2πi −1 3 √ + i 2 2 ! √ 6−2 √ 2π 2 2

Rational and Trigonometric functions over R line: Consider Z∞

f (x)eix dx

(3.3)

−∞

where f (x) is a real valued function of a real variable x. We shall assume that degree of denominator of f (x) exceeds degree of numerator of f (x) by atleast two and f (x) Z∞ has no poles on < line so that f (x)eix dx is convergent. −∞

Z∞ Note that real and imaginary part of

Z∞

ix

f (x)e dx are −∞

Z∞

ZR f (x)sin(x)dx. Consider

−∞

f (x)cos(x)dx and −∞

Z

ix

f (x)e dx where R > 0, we also consider −R

f (z)eiz dz

γR it

where γR is the semicircular R given |z| = R ⇒ z = Re where 0 ≤ t ≤ π. Consider the Jordan contour γ = [−R, R] ∪ γR where [−R, R] denotes the line segment from 98

−R to R. By Cauchy’s residue theorem, we have

ZR

Z

ix

f (z)eiz dz = 2πi(sum of residues of f (z)eiz at z = zi).

f (x)e dx + −R

γR

Jordan lemma Suppose R >Z 0 and further γR is the semicircular are given by z = Reit where 0 ≤ t ≤ π, then |eiz ||dz| < π. γR

Proof. Z

Zπ

iz

|e ||dz| = γR

it

|ei(Re ) ||iReit ||dt|

0

Zπ = R

it

|ei(Re ) ||eit ||dt|

0

Zπ = R

|e−R sin(t) ||eiR cos(t) ||dt|

0

Zπ = R

e−R sin(t) |dt|

0 π

Z2 = 2R

e−R sin(t) |dt|

0 π

Z Since t is real, then γR

|eiz ||dz| = 2R

Z2

e−R sin(t) dt. We have sin(t) ≥

0

99

2t π

if 0 ≤ t ≤ π2 ,

then π

π

Z2

Z2

e−R sin(t) dt ≤

0

2t

e−R π dt

0

" = Z

2t

e−R π   −R π2

|eiz ||dz| ≤ 2R

# π2 0

−π −R(1) [e − e0 ] 2R

γR

Note: It follows from Jordan lemma, we have Z f (z)eiz dz → 0 as R → ∞ so that γR

Z

f (z)eiz dz = 2πi

(sum of residues of f (z)eiz at z = zi)

−∞

where all the poles lie in the upper half plane. Z∞ 12.

cos(x) dx where a is real. x 2 + a2

0

Z∞ I = F (z) =

cos(x) dx ⇒ 2I = x 2 + a2

0 eiz z 2 +a2

Z∞

cos(x) dx. To evaluate this integral, note that x 2 + a2

∞

where simple poles at z = ±ia.

Consider Jordan contour, γ = [−R, R] ∪ γR where R > a

100

By Cauchy’s residue theorem we have, ZR

Z F (x)dx +

−R

F (z)dz = 2πi(Resz=ia F (z)) γR

Resz=ia F (z) = lim (z − ia) z→ia

Resz=ia F (z) = ZR

e−a 2ia 

Z F (x)dx +

−R

eiz (z − ia)(z + ia)

F (z)dz = 2πi

e−a 2ia

 =

πe−a a

(3.4)

γR

Note that Z Z 1 F (z)dz ≤ |eiz ||dz| 2 − a2 R γR

γR

π < → 0 as R → ∞ in (3.4) 2 R − a2 Z∞ F (x)dx =

πe−a as R → ∞ a

−∞

Z∞

1 cos(x) dx = 2 2 x +a 2



πe−a a



0

Z∞ 13.

xsin(x) dx where a is real. x 2 + a2

−∞

Z∞ I=

xsin(x) dx x 2 + a2

−∞

To evaluate this integral, note that F (z) =

zeiz z 2 +a2

where simple poles at z = ±ia.

Consider Jordan contour, γ = [−R, R] ∪ γR where R > a

101

By Cauchy’s residue theorem we have, ZR

Z F (x)dx +

−R

F (z)dz = 2πi(Resz=ia F (z)) γR

Resz=ia F (z) = lim (z − ia) z→ia

zeiz (z − ia)(z + ia)

e−a iae−a = 2ia 2

Resz=ia F (z) = Note that

Z Z 1 F (z)dz ≤ |z||eiz ||dz| → 0 as R → ∞ 2 2 R −a γR Z∞

γR

F (x)dx = 2πi(

ie−a ) 2

−∞

Z∞

xsin(x) dx = πe−a x 2 + a2

−∞

Z∞ 14.

sin(x) dx x

0

The function F (z) =

eiz z

has a simple pole at z = 0. Let 0 < r < R and let γ be

the closed curve in the figure. It follows from Cauchy’s theorem, Z F (z)dz = 0 γ

Breaking γ into its pieces

Z−r F (x)dx +

0= −R

ZR

Z

F (x)dx +

F (z)dz + γr

r

102

Z F (z)dz γR

(3.5)

where γr and γR are semicircles from −r to r and −R to R. But ZR

sin(x) dx = x

r

ZR

eix − e−ix 1 dx = 2ix 2i

r

=

1 2i 1 2i

eix 1 dx − x 2i

ZR

1 eix dx − x 2i

ZR

ZR

e−ix dx x

r Z−R

eiy dy y

−r

r

1 = 2i

eix − e−ix dx x

r

ZR r

=

ZR

eix 1 dx + x 2i

Z−r

eiy dy ⇒ 2i y

−R

r

ZR

sin(x) dx = x

r

ZR r

eix dx + x

Z−r

eiy dy y

−R

and now, Z iz Zπ iRcos(θ)−Rsin(θ) e e iθ dz = iRe z Reiθ 0

γR

When |eiRcos(θ) | = 1, we get Z iz Zπ e ≤ e−Rsin(θ) dθ dz z γR

0

By the method of Calculus we see that, for δ > 0 sufficiently small. The largest possible values of e−Rsin(θ) with δ ≤ θ ≤ π − δ is −Rsin(δ). This gives that Z iz Zπ−δ e dz ≤ 2δ + e−Rsin(θ) dθ. z γR

δ

 If  > 0 is given, choosing δ < 2 ther is an R0 such that e−Rsin(θ) < 3π for all R > R0 Z iz e   2  ⇒ dz ≤  + (π − 2δ) =  + (π − ) =  + z 3π 3π 2 3 γR Z iz e ⇒ lim dz = 0 R→∞ z γR

103

iz−1

Since e z has removable singularity at z = 0. There is a constant M > 0 such that eiz−1 z ≤ M for all |z| ≤ 1. Hence Z iz−1 Zπ iz−1 e e ≤ dz z |dz| ≤ M πr z 0

γr

Z ⇒ lim

r→0

eiz−1 dz = 0 z

(3.6)

and Z

Z dz = i dθ = −πi; z γr π Z Z Z iz dz e dz iz ⇒ lim e = lim = −πi(3.6) ⇒ dz = −πi r→0 r→0 z z z γr

γr

γr

Therefore (3.5) the equation becomes ZR 0 = 2i

sin(x) = −πi x

r

ZR ⇒

−π sin(x) = = x 2

r

Z∞

sin(x) x

0

as r → 0 and R → ∞. R∞ Type: Consider −∞ f (x) sin(x)dx. The argument implies the existence of integrals RR Rx lim f (x) sin(x)dx and not as lim lim −x2 1 f (x) sin(x)dx. However it turns out x1 →∞ x2 →∞

R→∞ −R

that this function does not pass difficulties since the function in question turns out to be even functions of x, so that for x1 , x2 > 0 we have Zx2

Zx1 f (x) sin(x)dx =

−x1

Z∞ Now cosider

Zx2 +

0

. 0

f (x)eix dx where f (x) is < valued function in < variable. Suppose

−∞

now that degree of denominator of f exceeds degree of numerator of f by 1 and that 104

f has no poles on R line . Rx2 f (x)eix dx where −x1 < 0 < x2 and consider the limit as To establish existence of −x1

x1 → ∞ and x2 → ∞. Clearly we cannot use semicircular arc. We will use a rectangular contour C : [−X1 , X2 ] ∪ [X2 , X2 + iY ] ∪ [−X1 + iY, X2 + iY ] ∪ [−X1 , X1 + iY ] where Y > 0, here [Z1 , Z2 ] where Z1 , Z2 ∈ C denote line segment from Z1 toZ2 . By cauchy residue theorem, we have Zx2

ix

Z

f (x)e dx + −x1

Z

iz

f (z)e dz +

[x2 ,x2 +iy]

Z

iz

f (z)e dz +

[x2 +iy,−x1 +iy]

f (z)eiz dz

[−x1 +iy,−x1 ]

= 2πi

X


Res f (z)eiz , zi (3.7)

Zi inC

where

P

is taken over all the poles of rectangular contour. When x1 , x2 and y

are large, then all poles of f (z)eiz in upper half plane are inside contour C. Under hypothesis, zf (z) is bounded. Then there exist M > 0 such that |zf (z)| ≤ M , where z ∈ C. Z

ZU

ix

Note that

f (z)e dz = i

Then, |f (x2 + iy)| <

[since z = x2 + iy]

0

[−x2 ,x2 +iy] M |x2 +iy|

f (x2 + iy)ei(x2 +iy) dy

≤

M x2

[since|z| > Rz]

Z M Zy f (z)eiz dz ≤ e−y dy ⇒ x2 [x2 ,x2 +iy] 0 Z Zy M ⇒ e−y dy f (z)eiz dz ≤ x2 [x2 ,x2 +iy] 0  −y y M e M M = = (1 − e−y ) < x2 −1 0 x2 x2 Z Similarly f (z)eiz dz < M . x1 [x1 +iy,x2 ] 105

Z

ZX2

iz

f (z)e dz = −

Now,

f (x + iy)ei (x + iy)dx

−X1

[x1 +iy,x2 +iy]

|f (x + iy)| <

M M ≤ |x + iy| y

Then, Z M ZX2 M −y e−y dx < f (z)eiz dz < e (x2 + x1 ). y y [x1 +iy,x2 +iy] −X1 From equation (3.7) X Z Z2 Z Z X ix iz f (x)e dx − 2πi Res(f (z)e , z ) ≤ + + i (x2 ,x2 +iy) (x2 +iy,−x1 +iy) (x1 +iy,x2 +iy) −X1 <

M M M + + e−y (x2 + x1 ) x2 x1 y

(3.8)

Note that left hand side of (3.8) above equation is independent of y for fixed x1 , x2 we have

M −y e (x2 y

+ x1 ) → 0 as y → ∞, therefore as x1 , x2 → ∞ ⇒ Right hand side

of(3.8) equation equal to zero Therefore 2πi

X

Res(f (z)eiz , zi ) =

z1 in C

ZX2

f (x)eix dx

−X1

Integrands with branch points Z∞ Consider xα f (x)dx where f (x) is real valued rational function in the real variable 0

x. We assume that degree of denominator exceeds degree of numerator by atleast two and that f (x) has no poles on positive real axis, and atmost a simple pole at the origin. We shall assume 0 < α < t. Here z α f (α) is not single valued. Now substituting x = U 2 and the integral become Z∞ 0

U 2α f (U 2 )2U.dU = 2

Z∞

U 2α+1 f (U 2 )dU =

Z∞ −∞

0

106

U 2α+1 f (U 2 )dU

Consider a new function F (z) = z 2α+1 f (z 2 ) for the function z 2α by choosing the branch so that arg(z 2α ) lies between −πα and 3πα. It is easy to see tat it is well defined and analytic in the region obtained from C by deleting the origin and the negative imaginary axis. It follows that, Jordan contour avoids this cut, then we can use Cauchy Residue theorem for f (z). Consider the Jordan Contour C, C : [−R, −δ] cupJ(δ) ∪ [δ, R] ∪ CR R>δ>0 −J(α) denotes the semicircular arc of the form z = δeit ; 0 ≤ t ≤ π By Cauchy residue theorem, we have Z−δ

Zδ

Z F (z)dz +

−R

F (z)dz +

Z F (z)dz +

R

J(α)

X

F (z)dz = 2πi

Res(F, zi )

(3.9)

zi in C

CR

whereZthe summation overZ all the poles of Jordan contour C. It is easily shown F (z)dz = 0 as δ → 0 and R → ∞. We get from the that F (z)dz = 0 and CR

J(α)

equation(3.9) Z∞

Z0 F (z)dz + −∞

Z∞ F (z)dz =

Res(F, zi )

zi in C

−∞

0

X

F (z)dz = 2πi

where the summation is taken over all poles of F (z) in upper half plane. On the other hand note that (−z)2α = (−1)2α z 2α = e2πiα z 2α Z∞ Z∞ Z∞ 2α+1 2 2α+1 2 f (z )dz = [z 2α+1 − (−z)2α+1 ]f (z 2 )dz ⇒ z f (z )dz = 2 z −∞

0

0

Z∞ =

z

2α+1

2

2πiα

Z∞

f (z )dz − e

0

z 2α+1 f (z 2 )dz

0

= (1 − e2πiα )

Z∞ 0

107

U 2α+1 f (U 2 )dU

since e2πiα 6= 1, we can get required value. Z∞ 1 x3 Example: dx 1 + x2 0 Z Z 1 z3 Consider dz = f (z)dz where C is the contour consisting of the large 1 + z2 C

C

semicircle on the upper half plane indented at the origin. Here we have to avoid the 1

branch point O of z 3 by indenting the origin. The only pole of f (z) in C is z = i. Then, 1

z3 ResF (z) = lim(z − i) z=i z=i (z + i)(z − i) 1

(cos π2 + i sin π2 ) 3 1 π π = = (cos + i sin ) 2i 6 6 √ 2i √ 1 −i 3 i ( + ) = (1 − 3i) = 2 2 2 4

Hence by Cauchy’s residue theorem, ZR

Z F (z)dz = C

F (z)dz +

R

F (z)dz +

Z F (z)dz +

−δ

CR

δ

= 2πi(

Now as R → ∞,

ZR

Z

J(δ)

π π π π 1 )(cos + i sin ) = π(cos + i sin ) 2i 6 6 6 6

F (z)dz → 0 and as δ → 0

CR

Z F (z)dz = 0 J(δ)

Then in (3.9) as R → ∞ and δ → 0 Z∞

Z∞ F (z)dz +

0

F (z)dz

F (z)dz =

−∞

108

π √ ( 3 + 1) 2

Since negative real axis is branch cut, Z∞

Z∞ f (x)dx +

0

f (xeiπ )(−dx) = π(cos

π π + i sin ) 6 6

0

Z∞

1 3

x dx + 1 + x2

0

Z∞

π

1

ei 3 x 3 π π = π(cos + i sin ) 2πi 2 1+e x 6 6

0

Equating real parts, Z∞

1

x3 π π (1 + cos )dx = πcos 2 1+x 3 6

0

Z∞ 0

√

1

π( 23 ) π cos( π6 ) x3 π = =√ dx = π 3 2 1+x (1 + cos 3 ) 3 2

Definition: 27 (Harmonic Functions). The real and imaginary parts of analytic functions are conjugate harmonic functions. All theorems of analytic functions are theorems on pair of harmonic functions. Definition: 28 (Definitions and Basic properties). A real valued function U (z) or U (x, y) defined and single valued in a region Ω is said to be harmonic (or) potential function in Ω,if it is continuous together with its partial derivatives of first 2 orders and satisfies the Laplace equation, (i.e)

∂2u ∂x2

+

∂2u ∂y 2

=0

Note: • Sum of harmonic functions and constant multiple of harmonic function are again harmonic. • A linear function of x and y is also harmonic (i.e, αx + βy) where α, β 6= 0 complex constants. The Laplace equation in polar co-ordinates:   ∂ ∂u ∂ 2 u r r + 2 =0 ∂r ∂r ∂θ 109

Results Transition from Harmonic to Analytic: Let U be harmonic in Ω,then f (z) = Now U =

∂u ,V ∂x

∂u ∂x

− i ∂u is analytic in Ω. ∂y

= − ∂u ∂y

⇒ f (z) = U + iV ∂U ∂ 2u ∂ 2u ∂V = =− 2 = 2 ∂x ∂x ∂y ∂y 2 ∂ u ∂V ∂U = =− and ∂y ∂x∂y ∂x

[Since

∂2u ∂x2

+

∂2u ∂y 2

= 0]

Hence U and V satisfies C-R equations and this partial derivatives Ux , Uy , Vx , Vy are all continuous. ∴ f (z) = U + iV = ux − iuy is analytic in Ω. Conjugate differential of du(∗du) : Let u(z) = u(x, y) be harmonic in Ω,then f (z) = ux − iuy analytic in Ω. f (z)dz = (ux − iuy )(dx + idy) = (ux dx + uy dy) + (ux dy − uy dx) We know that, du =

∂u dx + ∂u dy ∂x ∂y

(3.10)

If there exist a complex conjugate of u,ie, V (x, y)

in Ω. ⇒ f (z) = u + iv is analytic in Ω. ux = vy and uy = −vx Let us consider imaginary part of (3.10) ∂v ∂v dy + dx = dv ∂y ∂x ∴ (3.10) ⇒ f (z)dz = du + idv = d(u + iv) ux dy − uy dx =

But in general,there is no such single valued complex conjugate of u in Ω.In this case we can write ux dy − uy dx as ∗du. ∴ (3.10) becomes f (z)dz = du + i(∗du) where ∗du: conjugate differential of du. NOTE:

110

We know that v is a single valued complex conjugate of u,then ∗du = dv R Let γ be a cycle which is homologous to zero in Ω,then f dz = 0. Also γ

Z

f (z)dz = du + i ∗ du Z Z f (z)dz = du + i ∗du = 0.

γ

Since u is exact differential

γ

R

γ

du = 0

γ

Z ⇒

∗du = 0 γ

Theorem: 27. If u1 , u2 are harmonic in Ω then

R

u1 ∗du2 − u2 ∗du1 = 0 for every

γ

cycle γ which is homologous to zero in Ω. Proof. It is enough to prove that the integral vanishes in every rectangle in Ω by Cauchy’s theorem on rectangle (All rectangles are simply connected).Hence in γ both u1 and u2 have single valued harmonic conjugate v1 , v2 respectively. ∴ ∗du1 = dv1 and ∗du2 = dv2 u1 ∗ du2 − u2 ∗ du1 = u1 dv2 − u2 dv1 = u1 dv2 + v1 du2 − v1 du2 − u2 dv1 = (u1 dv2 + v1 du2 ) − (v1 du2 + u2 dv1 ) = (u1 dv2 + v1 du2 ) − d(u2 v1 )

(3.11)

(u1 + iv1 )d(v2 + iv2 ) = (u1 + iv1 )(dv2 ) = (u1 dv2 − v1 dv2 ) + i(v1 dv2 + v1 du2 ) Im((u1 + iv1 )d(v2 + iv2 )) = u1 dv2 + v1 du2

(3.12)

R R u1 ∗ du2 − u2 ∗ du1 = Im((u1 + iv1 )d(u2 + iv2 )) − d(u2 v1 ) γ γ γ R Since d(u2 v1 ) is exact differential ⇒ d(u2 v1 ) = 0 sub (3.11 ) in (3.12 )

R

γ

By Cauchy’s theorem,in general form, Z Im[f1 ∗ f2 ]dl = 0 where f1 = u1 + iv1 γ

111

and

f2 = u2 + iv2

∴

R

u1 ∗ du2 − u2 ∗ du1 = 0 for every cycle γ which is homologous to zero in Ω.

γ

Theorem: 28 (Mean Value Property For Harmonic Functions:). The arithmatic mean of the harmonic function over a concentric circle |z| = r is a linear function of R 1 log r. 2π udθ = α log r + β and if u is harmonic in a disc α = 0 and the arithmetic |z|

mean is constant. Proof. We know that Z u1 ∗ du2 − u2 ∗ du1 = 0

(i)

(3.13)

γ

Put u1 = log r and u2 = u in (3.13) Z ⇒ log r ∗ du − u ∗ d(log r) = 0

(3.14)

γ

We know that log r and u are harmonic.Take a cycle γ in the following region 0 < |z| < ρ.Let c1 be circle with |z| = r1 and c2 be circle with |z| = r2 and let r1 < r2 and γ be a cycle homologous to zero in Γ,then (3.14 ) holds. Then taking c2 − c1 = γ Z ⇒

log r ∗ du − u ∗ d(log r) = 0

(3.15)

c2 −c1

dθ (If γ is a regular curve with equation z(t) = x(t) + iy(t)).The direction ∗du = r ∂u ∂r of tangent is determined by the angle α = argz 0 (t) and we can write dz = ireiθ dθ dx = |dz| cos α,

dy = |dz| sin α.

The normal which points to the right of direction of tangent β = α −

π 2

and thus

cos α = − sin β and sin α = cos β. ∴ Expression

∂u ∂n

=

∂u ∂x

cos β +

∂u ∂y

sin β is the directional derivative of u along

the normal, (right hand normal derivative with respect to curve γ) then, ∗du =

112

∂u ( ∂n )|dz| = r ∂u dθ] ∴ (3.15) becomes: ∂r   Z ∂u ∂(log r) log r r dθ − ur dθ ∂r ∂r c2 −c1 Z 1 ∂u r log r dθ − ur( )dθ ∂r r c2 −c1 Z ∂u r log r dθ ∂r c2 −c1 Z Z ∂u ∂u r log r dθ − r log r dθ ∂r ∂r c2 c1  Z  ∂u r log r − u dθ ∂r

= 0 = 0 Z udθ

= c2 −c1

Z

Z udθ −

= c2

udθ c

1  Z  ∂u r log r dθ − udθ = ∂r

c2

c1

 Z  Z ∂u r log r dθ − udθ = constant ∂r |z|=r |z|=r Z Z ∂u udθ = r log r dθ + constant ∂r |z|=r |z|=r Z Z c 1 1 ∂u udθ = rlogr dθ + 2π 2π ∂r 2π |z|=r

1 2π

|z|=r

Z udθ = α log r + β |z|=r

c r ∂u dθ;β = 2π ∂r R (ii) We know that rdu=0

where α =

1 2π

R

γ

Z ⇒ γ

1 ∂u r dθ = ∂r 2π

Z r

∂u dθ = 0 ∂r

γ

∴ α = 0.

sub in (3.16) 113

(3.16)

1 2π

Z udθ = β |z|=r

Note: Changing the centre of circle from origin to z0 , we have (z − z0 ) = reiθ ⇒ z = z0 + reiθ Z2π Z2π 1 1 U (z)dθ = U (z0 )dθ = U (z0 ) 2π 2π 0

0

U (z0 ) =

1 2π

Z2π

U (z0 + reiθ )dθ

0

The value of U(z) at z = z0 is the arithmetic mean value of centre of circumference. Theorem: 29 (MAXIMUM PRINCIPLE FOR HARMONIC FUNCTIONS). A non constant harmonic function has neither a maximum nor a minimum in its region of definition. Consequently the maximum and minimum on a closed, bounded set E are taken on the boundary of E. Proof. We know that 1 U (z0 ) = 2π

Z2π

U (z0 + reiθ )dθ

0

The value of harmonic function at centre of the circle |z − z0 | = r is contained in the region of definition of Ω. It is equal to arithmetic mean of its values of the circle subject to condition that the closed disc |z − z0 | ≤ r is contained in Ω. If possible, let U(z) attains maximum in interior of Ω say at z0 . ⇒ U (z) ≤ U (z0 ) whenever |z − z0 | = r. The strict inequality is true for a single value of θ U (z0 + reiθ ) < U (z0 )

114

then by continuity, it would hold for entire boundary circle contained in neighbourhood at z0 . 1 U (z0 ) = 2π

Z2π

1 U (z0 + reiθ )dθ < 2π

0

Z2π U (z0 )dθ = U (z0 ) 0

which is a contradiction. ∴ U(z) is constant in neighbourhood of z0 . Extending at successive points in the same procedure, it attains the minimum in interior of Ω. Then -U (z0 ) is maximum for -U(z). U (z0 ) is a constant. By maximum modulus theorem, in a compact set, the harmonic function has maximum / minimum on boundary. POISSON THEOREM: Let us find a bilinear transformation |z| = R and |ζ| ≤ 1. Let the points z=a on |z| = R is mapped onto centre of |ζ| = 0. Also inverse points mapped onto inverse points (i.e.,) inverse transformation |ζ| ≤ 1 mapped onto |z| = R where ζ =0 is mapped to z=0. Let U(z) be harmonic in |z| ≤ R. Then U(S(ζ)) is harmonic in |ζ| ≤ 1. By mean value property, 1 U (a) = 2π

Z2π U (z)dθ 0

1 U (S(ζ)) = 2π

Z2π

1 U (S(ζ))d(argζ) = 2π

0

Z U (S(ζ))d(argζ) |ζ|=1

ζ = eiθ ⇒ dζ = ieiθ dθ ⇒

dζ = dθ = d(argζ) iζ

115

(3.17)

Let us see the transformation which maps |ζ| ≤ 1 to |z| ≤ R and ζ=

R(z − a) R2 − a ¯z

(3.18)

Taking log, logζ = logR + log(z − a) − log(R2 − a ¯z) dζ ζ

dz a ¯dz + 2 z−a R −a ¯z 1 a ¯ = dz( + ) z − a R2 − a ¯z = 0+

Put R2 = z z¯ dζ ζ

1 a ¯ dz + ) z − a z(¯ z−a ¯) iz a ¯ z + )dθ d(argζ) = ( z − a z¯ − a ¯ ⇒

= (

(3.19)

where (arg z = θ) a ¯ z(¯ z−a ¯) + a ¯(z − a) z + = z − a z¯ − a ¯ |z − a|2 z z¯ − a ¯z + a ¯z − a¯ a = 2 |z − a| 2 R − |a|2 = |z − a|2

z z¯ = R2

Also, z a ¯ 1 z a ¯ z¯ a + = [( + )+( + )] z − a z¯ − a ¯ 2 z − a z¯ − a ¯ z¯ − a ¯ z−a 1 z + a z¯ + a ¯ = [ + ] 2 z − a z¯ − a ¯ z+a = Re[ ] z−a ∴ (3.19) becomes d(argζ) = Re(

z+a R2 − |a|2 )dθ = dθ z−a |z − a|2 116

(3.17) becomes 1 U (a) = 2π

Z U (S(ζ))d(argζ) |ζ|=1

1 = 2π

Z

R2 − |a|2 U (S(ζ))dθ |z − a|2

|ζ|=1

1 = 2π

Z Re(

z+a )U (S(ζ))dθ z−a

|ζ|=1

R2 − |a|2 U (z)dθ |z − a|2

Z

1 U (a) = 2π

z = S(ζ) = Reiθ

|ζ|=1

It is called Poisson integral formula. Express Poisson Integral formula in Polar Co-ordinates: We have Z 1 R2 − |a|2 U (a) = U (z)dθ 2π |z − a|2

(3.20)

|ζ|=1

where z = Reiθ Setting a = reiθ and z = Reiθ ⇒ |z − a|2 = |Reiθ − reiθ |2 = |R(cos θ + i sin θ) − r(cos φ + i sin φ)|2 = |(R cos θ − r cos φ) + i(R sin θ − r sin φ)|2 = (R cos θ − r cos φ)2 + (R sin θ − r sin φ)2 = R2 cos2 θ − 2Rr cos θ cos φ + r2 cos2 φ + R2 sin2 θ + r2 sin2 φ − 2Rr sin θ sin φ = R2 + r2 − 2Rr(cos θ cos φ + sin θ sin φ) = R2 + r2 − 2Rr(cos(θ − φ)) and |a|2 = r2 sub in (3.20), 1 u(re ) = 2π iθ

Z2π

(R2 − r2 )U (Reiθ ) dθ, R2 + r2 − 2Rr cos(θ − φ)

0

This is the Poisson integral formula in polar co-ordinates. NOTE: 117

r
We have Z

1 U (a) = 2π

Re(

z+a )U (S(ζ))dθ z−a

|ζ|=1

Z

1 = 2π

Re(

z+a )U (z)dθ z−a

|z|=R

Put a=z and z = ζ 1 ⇒ U (z) = 2π

Z Re(

dζ ζ +z )U (ζ) ζ −z iζ

where

ζ = eiθ

|ζ|=R

Theorem: 30 (Schwarz Formula). Suppose that U(z) is harmonic for |z| < R, continuous for |z| ≤ R then Z 1 R2 − |a|2 U (a) = U (z)dθ, 2π |z − a|2

∀|a| < R

|z|=R

Proof. we know that Z Z 1 R2 − |a|2 1 u(a) = |z| = R u(z)dz = 2π |z − a|2 2π



 z+a Re u(z)dθ z−a

|z|=R

put a = z and z = r = eiθ , dr = ieiθ dθ = irdθ    Z  r+z dr 1 u(r) ⇒ u(z) = Re 2πi r−z r |r|=R

The exp in breacket is analytic for |z| < R. It follows that u(z) is real part of  Z  r+z 1 dr f (z) = u(r) + iC 2πi r−z r |r|=R

where C arbitrary real constant. This formula is called Schwarz formula. Note If u = 1, we get Z

R2 − |a|2 dθ = 2π |z − a|2

|z|=R

118

Chapter 4

SERIES AND PRODUCT DEVELOPMENT (Power series expansion) WIESTRASS THEOREM FOR UNIFORM CONVERGENCE: Consider a sequence fn (z) where each fn (z) is defined and analytic in region Ωn . The limit function f (z) must also be considered in some region Ω and clearly if f (z) is defined to be in Ω, each point of Ω belongs to all Ωn for n ≥ n0 (i.e) Given  > 0 ∃ n0 > 0 3 |fn (z) − f (z)| <  whenever n ≥ n0 and z ∈ Ω. Infact in the most typical case where regions Ωn form a increasing sequence S Ω1 ∈ Ω2 ∈ . . . ∈ Ωn and Ω = Ωn . In these circumstances no single function n

defined in all of Ω. Yet limit f (z) may exists at all points of Ω. For eg: fn (z) =

z 2z 2 +1

and let Ωn be disc |z| <

1 1

2n

then it is evident that lim fn (z) = z n→∞

in the disc |z| < 1. we form the difference z −2z n+1 fn (z) − z = n −z = n 2z + 1 2z + 1 For any given value of z, we can make log  |z|n < by taking n > 4 log

4 
1 |z|




if  < 1

we have 2( 4 ) 2|z|n+1 < = 2|z|n + 1 2( 4 ) + 1 lim fn (z) = z

|fn (z) − z| = n→∞

119

 2  2

+1

=

 < +2

Theorem: 31 (Wiestrass theorem for uniform convergence:). Suppose that fn (z) is analytic in region Ωn and that the sequence fn (z) converges to a limit function f (z) in a region Ω uniformly on every compact subset of Ω, then f (z) is analytic in Ω. 0

0

Moreover fn (z) converges uniformly to F (z) on every compact subset. Proof. Let a any point in Ω and |z − a| ≤ r be a closed disc contained in Ω. This assumption implies that this disc lies in Ωn because of region Ωn form an open covering for |z − a| ≤ r. Since this disc is compact then it has a finite subcover. this means that it is contained in a finite Ωn0 for n = n0 . Let γ be any closed curve contained in |z − a| < r. Then by Cauchy theorem: Z fn (z)dz = 0 for n = n0 γ

because of uniform of convergence on γ, Z lim fn (z)dz = 0 n→∞

γ

and since Z fn → f ⇒

f (z)dz = 0 γ

∴ By Morera’s theorem: f (z) is analytic in |z − a| < r (i.e) analytic in a circular neighbourhood of a. Since a is arbitrary ⇒ f (z) is analytic in Ω. By Cauchy Integral formula: 1 fn (z) = 2πi

Z C

120

fn (ζ)dζ ζ −z

C : circle with |ζ − a| ≤ ρ, then 1 lim fn (z) = n→∞ 2πi lim fn (z) =

n→∞

fn0 (z) = 0

lim fn (z) =

n→∞

=

1 2πi 1 2πi 1 2πi 1 2πi

Z

fn (ζ)dζ n→∞ ζ − z lim

C Z

C Z

C Z

f (ζ)dζ = f (z) ζ −z f (ζ)dζ (ζ − z)2 fn (ζ)dζ n→∞ (ζ − z)2 lim

C Z

f (ζ)dζ = f 0 (z) (ζ − z)2

C

This convergence is uniform in |ζ − a| < ρ ≤ r. Hence any compact subset of Ω can be covered by finite number of such discs. Therefore the convergence is uniform on every subset of Ω. NOTE: If a series with analytic terms converges uniformly on every compact subset of Ω. Then sum of f (z) is analytic in Ω and the series can be differentiated term by term. Proof. Let us take simple closed curve γ in |z − a| ≤ r. Now f (z) = f1 (z) + f2 (z) + . . . + fn (z) To prove, 0

0

0

f (z) = f1 (z) + . . . + fn (z), (4.1) becomes, 1 f1 (z) 1 fn (z) 1 f (z) = + ... + . 2 2 2πi (z − a) 2πi (z − a) 2πi (z − a)2 Taking integral over γ Z Z Z 1 f (z) 1 f1 (z) 1 fn (z) = + ... + 2 2 2πi (z − a) 2πi (z − a) 2πi (z − a)2 γ

γ

γ

121

(4.1)

0

0

0

⇒ f (a) = f1 (a) + . . . + fn (z) 0

0

0

since a is arbitrary ⇒ f (z) = f1 (z) + . . . + fn (z) Theorem: 32 (Hurwitz Theorem:). If the functions fn (z) are analytic and not equal to zero in a region Ω and if fn converges to f (z) on every compact subset of Ω, then f (z) is either = 0 or never to zero in Ω. Proof. Suppose that f (z) 6= 0 then we have to show that f (z) is never zero in Ω. Suppose f (z) has a zero in Ω. Since zeros of analytic functions are isolated, ∃r > 0 ∈ R, such that f (z) is defined in 0 < |z −z0 | < r, where f (z0 ) = 0 in particulars f (z) 6= 0 on the circle |z − z0 | < r and denote this circle as Γ . Let  = min |f (z)| : z ∈ Γ so that  ≤ |f (z)| for every z ∈ Γ. By Hypothesies fn (z) ⇒ f (z) on Ω ⇒ |fn (z) − f (z)| <  ≤ |f (z)| ⇒ |fn (z) − f (z)| < |f (z)| Then by Rouche’s theorem, f (z) and fn (z) − f (z) + f (z) have same no.of zeros in Γ f (z) and fn (z) − f (z) + f (z) have same number of zeros in Γ but f (z) has a zero at z = z0 ⇒ fn (z) must have a zero at z = z0 which is contraction to our hypothesis as fn (z) 6= 0 ∴ f (z) is never zero in Γ ⊂ Ω ∴ f (z) is never zero in Ω. Taylor series: Theorem: 33. If f (z) is analytic in a region Ω containing z0 then the representa0

tion, f (z) = f (z0 ) + (z − z0 )f (z0 ) +

(z−z0 )2 00 f (z0 ) 2!

of center z0 contained in Ω.

122

+ . . . is valid in the largest open

Note: 1 The radius of convergence of Taylor’s series is atleast equal to the shortest distance from z0 to boundary of Ω. It may be larger. But if it is, there is no gaurenty that series still represent f (z), at all points in Ω and circle of convergence. We know that series development z2 + ... 2! z4 + + ... 4! z5 + − ... 5!

ez = 1 + z + z2 cos z = 1 − 2! z3 sin z = 1 − 3!

We know that every convergent power series is its own Taylor series and also power series can be differentiated term by term which is a consequence of weierstrass theorem now we want to represent a fractional power of z. For this case we have to choose a well defined branch and we have to choose a center z0 6= 0. It is enough to develop the function log(1 + z) about origin. Since this branch is analytic in |z| < 1 and the radius of convergence is atleast 1. (1 + z)µ = 1 + µz + µC2 z 2 + µC3 z 3 + . . . + µCn z n where binomial coefficients are defined: µCn = µCn =

µ(µ−1)(µ−2)...(µ−n−1) n(n−1)...3.2.1

µ! n!(µ−n)! 2

3

and log(1 + z) = z − z2 + z3 + . . . if the logarathmic series

have a radius of convergence > 1, These log(1 + z) is bounded in |z| > 1. Since this is not the case radius of convergence=1. Similarly binomial series are convergent in Therefore function (1 + z)µ and all its derivatives would be bounded in |z| < 1. Thus radius of convergent is precevely 1, except in trivial case, the binomial series reduces a polynomial. Consider the expansion:

1 = (1 + z 2 )−1 = 1 − z 2 + z 4 − z 6 + . . . 1 + z2 123

integrating Z

Z 1 dz = (1 − z 2 + z 4 − z 6 + . . .)dz 2 1+z z3 z5 z7 + − + . . . = arc tan z tan−1 z = z − 3 5 7 R dz where the branch is uniquely determined as arc tan z = 1+z 2 any path inside unit circle −1 1 1.3 4 1.3.5 6 1 z + z + ... = (1 − z 2 ) 2 = 1 + z 2 + 2 2 2.4 2.4.6 1−z Z 1 1 3 1.3 z 5 1.3.5 z 7 √ z + + + ... = z+ 2.3 2.4 5 2.4.6 7 1 − z2

√

this represent the principal branch of arc(sin z) with a real part between

−π 2

and

π 2

Note: Now we introduce the notation [z n ] for any functions which analytic and has a zero of atleast order n at origin. It denotes the function which contains a factor z n , with this notation any functions analytic at origin can be written in the form f (z) = a0 + a1 z + a2 z 2 + . . . + an z n + [z n+1 ] where the coefficients are uniquely determined and equal to taylor coefficients of f (z). Thus in order to find the first n coefficient of f (z) in taylor series it is sufficient to determined Pn (z) 3 f (z) − Pn (z) has a zero of order of atleast n + 1 at origin. Result: f (z) = a0 + a1 z + a2 z 2 + . . . + an z n + . . .

and

g(z) = b0 + b1 z + b2 z 2 + . . . + bn z n + . . . ⇒ f (z) = Pn (z) + [z n+1 ]

and

g(z) = Qn (z) + [z n+1 ] f (z) + g(z) = Pn (z) + Qn (z) + [z n+1 ] f (z).g(z) = a0 b0 + (a0 b1 + a1 b0 )z + . . . + (a0 bn + a1 bn−1 + . . . + an−1 b1 + an b0 )z n +[z n+1 ]

124

It is clear that f (z)g(z) = Pn (z)Qn (z) + [z n+1 ] f (z) = Rn (z) + [z n+1 ] g(z) The coefficient of Rn (z) are Taylor coefficients of

f (z) . g(z)

Now we develop compo-

sition f (g(z)). In this case if g(z) is developed around z0 the expansion of f (w) in power of w − g(z0 ). Assume z0 = 0 ⇒ g(0) = 0 Therefore we can set f (w) = a0 + a1 w + . . . + an wn + . . .

and

g(z) = b0 + b1 z + . . . + bn z n + . . . Using the same notation f (w) = Pn (w) + [wn+1 ] and g(z) = Qn (z) + [z n+1 ] with Qn (0) = 0 subs w = g(z). We observe f (g(z)) = Pn (Qn (z))+[z n+1 ] and the Taylor’s coefficients of f (g(z)) are those of Pn (Qn (z)). Now we develop inverse function of an analytic function w = g(z). Suppose g(0) = 0 we are looking for branch of inverse function z = g −1 (w) which is analytic in neighbourhood of origin and vanishes for w = 0 (i.e) g −1 (0) = 0. The existence of inverse function has the necessary and sufficient condition that the given map is a necessary and sufficient condition that the given map is a conformal mapping (i.e) g −1 (0) = 0. We assume that g(z) = a1 z + a2 z 2 + . . . + an z n + [z n+1 ] with a1 6= 0 we have to find Pn (w) such that Pn (Qn (z)) = z + [z n+1 ] under a1 6= 0, the notations [z n+1 ] and wn+1 are interchangeable. z = Pn (Qn (z)) + [z n+1 ] = Pn (w) + [wn+1 ]. Hence Pn (w) determines the coefficients of g −1 (w)

125

Examble: z3 z5 z7 + − + ... 3 5 7 z3 z5 w+ − + [z 7 ] 3 5 z3 z5 1 z3 z5 1 − + [z 7 ])3 − (w + − + [z 7 ])5 + [z 7 ] w + (w + 3 3 5 5 3 5 1 1 1 w + w3 + (w2 )(w + [w3 ])3 − w5 + [w7 ] 3 3 5 1 3 2 5 w + w + w + [w7 ] 3 15

w = arc tanz = z − ⇒z = z = z = =

Thus the development of tan z.

Theorem: 34 (Laurent’s theorem:). Let f (z) be analytic in the ring shaped region D bounded by a concentric circle C1 and C2 with center z0 and radii ρ1 and ρ2 with ρ1 > ρ2 . Let z be any point of D, then f (z) =

∞ X

n

An (z − z0 ) +

n=0

where An =

1 2πi

R C1

f (ζ)dζ (ζ−z0 )n+1

∞ X

Bn (z − z0 )−n

n=0

and Bn =

1 2πi

R C1

f (ζ)dζ . (ζ−z0 )−n+1

Proof. If z ∈ D then by cauchy integral formula for doubly connected region, we have: 1 f (z) = 2πi

Z

f (ζ)dζ − ζ −z

C1

Z

f (ζ)dζ ζ −z



C2

we consider 1st integral in (4.2) for any point ζ on C1 , we consider 1 1 1  = = ζ −z ζ − z0 − (z − z0 ) (ζ − z0 )

1 1−

z−z0 ζ−z0



 −1 1 z − z0 = 1− (ζ − z0 ) ζ − z0   1 z − z0 z − z0 2 z − z0 n 1 = 1+ +( ) + ... + ( ) 0 ζ − z0 ζ − z0 ζ − z0 ζ − z0 (1 − z−z ) ζ−z0 =

1 (z − z0 ) (z − z0 )2 (z − z0 )n 1 + + ( ) + . . . + ζ − z0 (ζ − z0 )2 (ζ − z0 )3 (ζ − z0 )n (ζ − z) 126

(4.2)

Multyplying

1 2πi

R

f (ζ)dζ on both sides

C1

1 2πi

Z

Z f (ζ)dζ z − z0 f (ζ)dζ + + ... ζ − z0 2πi (ζ − z0 )2 C1 C1 n Z f (ζ)dζ (z − z0 ) + 2πi (ζ − z0 )n (ζ − z)

f (ζ)dζ 1 = ζ −z 2πi

C1

Z

(4.3)

C1

Consider Rn =

(z−z0 )n 2πi

R C1

f (ζ)dζ (ζ−z0 )n (ζ−z)

and writing An =

1 2πi

R C1

f (ζ)dζ . (ζ−z0 )n+1

Then equation (4.3 ) becomes Z 1 f (ζ)dζ = A0 + (z − z0 )A1 + (z − z0 )2 A2 + ... + (z − z0 )n−1 An−1 + Rn . (4.4) 2πi ζ −z C1

We now show that Rn → 0 as n → ∞ Let |z − z0 | = r so that ρ2 < r < ρ1 . Also |ζ − z0 | = ρ1 . Then |ζ − z| = |ζ − z + z0 − z0 | = |(ζ − z0 ) − (z − z0 )| ≥ ρ1 − r then |z − z0 |n |Rn | ≤ 2π

|f (ζ)||dζ| |ζ − z0 |n |ζ − z| C1 Z |z − z0 |n M1 < |dζ| 2π ρn1 (ρ1 − r) Z

C1

n

r M1 (2πρ1 ) n 2π ρ1 (ρ1 − r)  n r M1 ρ1 = ρ1 (ρ1 − r) <

ζ ρ1

|Rn | → 0 as n → ∞.(Since ( ρr1 ) → 0 Since

< 1).

From equation (4.4) we have 1 2πi

Z C1

∞

f (ζ)dζ X = An (z − z0 )n ζ −z n=0

127

(4.5)

We now consider 2n d integral in (4.2). For any point ζ on C2 we consider: −1 1 = ζ −z (z − z0 ) − (ζ − z0 )  −1 1 ζ − z0 = 1− (z − z0 ) z − z0   1 ζ − z0 ζ − z0 2 ζ − z0 n 1 1+ +( ) + ... + ( ) = ζ−z0 z − z0 z − z0 z − z0 z − z0 (1 − z−z ) 0

1 (ζ − z0 ) (ζ − z0 )2 (ζ − z0 )n 1 = + + ( ) + . . . + z − z0 (z − z0 )2 (z − z0 )3 (z − z0 )n (z − ζ) 1 2πi

Z

Z f (ζ)dζ 1 f (ζ)dζ + (ζ − z0 ) + . . . z − z0 2πi (z − z0 )2 C2 C2 Z 1 f (ζ)dζ (ζ − z0 )n−1 + Sn + 2πi (z − z0 )n

f (ζ)dζ 1 = ζ −z 2πi

C2

Z

(4.6)

C2

where 1 Sn = 2πi

Z

1 (ζ − z0 )n f (ζ)dζ = n z − ζ (z − z0 ) 2πi(z − z0 )n

C2

Z

(ζ − z0 )n f (ζ)dζ (ζ − z

C2

To show that Sn → 0 as n → ∞. We observe, |ζ − Z0 | = ρ2 and |z − z0 | = r so that |z − ζ| = |(z − z0 ) − (ζ − z0 )| ≥ r − ρ2 and since f is bounded there exists M2 such that |f | < M2 on C2 and Z 1 1 |ζ − z0 |n |Sn | ≤ ||f (ζ)||dζ| 2π |z − z0 |n |ζ − z| C2 M2 (ρ2 )n

1 1 (2πρ2 ) 2π rn (r − ρ2 ) M2 ρ2  ρ2 n = r − ρ2 r R f (ζ)dζ 1 |Sn | → 0 as n → ∞ and Substitute Bn = 2πi (ζ−z0 )−n+1 <

C2

−1 2πi

Z

f (ζ)dζ = B1 + B2 (z − z0 )−2 + . . . (ζ − z0 )

C2

=

∞ X n=1

128

Bn (z − z0 )−n

(4.7)

sub (4.6) and (4.7) in (4.2) f (z) =

∞ X

n

An (z − z0 ) +

n=0

∞ X

Bn (z − z0 )−n

(4.8)

n=1

Note: We observe that Bn = A−n . Hence if C is any circle of radius ρ with center z0 such that ρ2 < ρ < ρ1 , the integral is analytic in the annulus ρ2 < |ζ − z0 | < ρ1 . We see that 1 An = 2πi

Z

f (ζ)dζ 1 ; B = n (ζ − z0 )n+1 2πi

C

Z

f (ζ)dζ . (ζ − z0 )−n+1

C2

Hence the series (4.8) can be written as ∞ X

f (z) =

An (z − z0 )n .

n=−∞

Theorem: 35. Prove that Laurent series is unique. Proof. Suppose we have obtained in any manner or as definition of f (z), the formula ∞ P f (z) = An (z − z0 )n is the Laurent’s development. If possible we can have n=−∞

another Laurent’s development as: f (z) =

∞ X

Pn (z − z0 )n

To prove:An = Pn

n=−∞

Let C be circle |z − z0 | = ρ where ρ2 < ρ < ρ1 then coefficient An is: ∞ R f (ζ)dζ P 1 and substitute f (ζ) = Pm (ζ − z0 )m dζ An = 2πi n+1 (ζ−z0 ) m=−∞

C

⇒ An

1 = 2πi

Z C

An

∞ X 1 Pm (ζ − z0 )m dζ (ζ − z0 )n+1 m=−∞

Z ∞ 1 X (ζ − z0 )m = Pm dζ 2πi m=−∞ (ζ − z0 )n+1 C

129

put ζ − z0 = ρeiθ ⇒ dζ = ρieiθ dθ 2π

An

Z ∞ 1 X Pm (iρeiθ )m−n−1 iρiθ dθ = 2πi m=−∞ 0

=

1 2π

∞ X

Z2π Pm

n=−∞

ρm−n ei(m−n)θ dθ

0

2π

Z ∞ 1 X m−n Pm ρ ei(m−n)θ dθ ∴ An = 2π n=−∞ 0

Taking I =

1 2π

R2π

ei(m−n)θ dθ

0

Case:(i) m 6= n : 1 I = 2π

Z2π

ei(m−n)θ dθ

0

π 1 ei(m−n)θ = 2π i(m − n) 0 1 1 = (1 − 1) 2πi m − n 

Case:(ii) m = n : 1 I= 2π

Z2π

e0 dθ = 1

0

substitute m = n in (4.9 ) An

∞ X

Pn

⇒ A n = Pn

n=−∞

∴ It is unique. 1. Differentiating (1 − 2αz + z 2 ) 2α−2z p1 (α) = = α. 3 2 2(1−2αz+z ) 2 z=0 solution:

−1 2

with respect to z we obtain

130

(4.9)

To compute higher order Legendre polynomial we differentiate (1 − 2αz + z 2 )

−1 2

and

its Taylor’s series to obtain α−z (1 − 2αz +

=

3

z2) 2

nPn (α)z n−1

n=1

α−z p

∞ X

2

(1 − 2αz + z 2 )

= (1 − 2αz + z )

∞ X

nPn (α)z n−1

n=1

Hence ∞ X

n

nPn (α)z −

n=1

∞ X

Pn (α)z

n−1

n=1

=

∞ X

nPn (α)z

n=1

n−1

−

∞ X

n

2αnPn (α)z +

n=1

∞ X

nPn (α)z n+1

n=1

Invoking elementary limit properties and using the fact that a f unc = 0. iff all its Taylor coefficients= 0. We equate terms to obtain the reference αPn+1 (α) − Pn (α) = (n + 2)Pn+2 (α) − 2α(n + 1)Pn+1 (α) + nPn (α) i 1 h Pn+2 (α) = (2n + 3)αPn+1 (α) − (n + 1)Pn (α) n+2 so P2 (α) = P3 (α) = = P4 (α) = =

1 (3α2 − 1) 2 i 1 h 5α (3α2 − 1) − 2α 3 2 i 1h 3 5α − 3α 2 i 1 h 7α 3 (5α3 − 3α) − (3a2 − 1) 4 2 2 i 1h 35α4 − 30α2 + 3 4

EX:3 2. Observe that

sin(z) z

=

1 z

∞ P n=0

(−1)2 2n+1 z (2n+1)!

soln:

∞

1 X (−1)2 2n sin(z) = z z z n=0 (2n + 1)! 131

so

sin z z

6= 0 in some open disc about z = 0. Hence the function z → log sinz z is

homomorphic in an open disc about z = 0. By taking the principal branch of logarithm and substituting  sinz  sinz = log 1 − ( ) log z z  m ∞ P (−1)n 2n 1 − z ∞ (2n+1)! X n=0 = m m=1 ∞  2 X
m  1  z Z 4 z6 = − − + − [z]8 3! 5! 7! m m=1 Set P (z) =

z2 3!

−

z4 . 5!

Then, h z 6 P (z) + [z]8 p(z)2 + [z]8 P (z)3 + [z]8 i sin z log( ) = − + + + z 7! 1 2 3 h z2 z4 z6 1  z4 i 6 6 2z z 8 = − − + + − + + [z] 3! 5! 7! 2 3! 3!5! 3(3!)3 z4 z6 z2 − + [z 8 ]. = − − 6 180 2835

3. Page 186(4) Solution: Put f (z) =

1 .Then ez −1

it has poles at z = 0, ±i2π, ±i4π, ... Thus f (z) is analytic in

the annulus 0 < |z| < 2π. Hence it and its an Laurent’s expansion in this annulus. We can write ∞ X

1 f (z) = An z where An = 2πi n=−∞ n

Z |z|=1

f (z) dz z n+1

we have 1 An = 2π

Z |z|=1

z −n−1 dz ez − 1

For n ≤ −2, the integral has singularity (removable) at z = 0 and no other singularity in unit circle. Thus An = 0(By Cauchy integral formula). for n = −1 Z 1 dz A1 = z 2π |z|=1 e − 1  1   z  = Res z = z =1 z=0 e − 1 e − 1 z=0 132

for n ≥ 0, z = 0 is a pole of order n + 2 of function

z −n−1 ez −1

Therefore, 1 dn+1  n+2 z −n−1  z z=0 ez − 1 (n + 1)! dz n+1 ez − 1 z=0 1 dn+1  z  = (n + 1)! dz n+1 ez − 1 z=0

An = Res

 z −n−1 

=

put  h(z) =

z ez −1

0

for z ∈ C − {0} for z = 0

Then h(z) is an entire function we have An =

hn+1 (0) (n+1)!

for n ≥ 0.

h(z) admits a Taylor’s eapansion,

h(z) = a0 + a1 z + a2 z 2 + · · · with an =

h(n)(0) , i.e. An = an+1 n!

for n ≥ 0. Thus we need to find a1 , a2 , a3 , a4 which corresponds to A1 , A2 , A3 . We Have a0 + a1 z + a2 z 2 + · · · = h(z) =

z z = = 2 ez − 1 (z + z2! + · · · ) 1+

z 2!

1 2 + z3! + · · ·

‘ 

1+

  z z2 z3 z4 + + + + [z 5 ] a0 + a1 z + · · · + a4 z 4 + [z 4 ] = 1 2! 3! 4! 5!

Equating constants, a0 = 1 Equating coefficients, z ⇒

a0 2!

+

a1 1!

= 0 ⇒ a1 = − 21

a0 a1 z2 ⇒ + + a2 = 0 3! 2! ha i h a1 1 1i 1 0 a2 = − + =− − = 6 2 6 4 12 h(−z) =

−z e− z−1

=

−zez 1−ez

=

zez . 1−ez

Thus

h(−z) − h(z) =

zez z − =z ez − 1 ez − 1 133

(4.10)

Thus z = (a0 − a1 z + a2 z 2 + · · · ) − (a0 + a1 z + a2 z 2 + · · · ) = −2[a1 z + a3 z 3 + · · · ] Thus a2k+1 = 0 for ∀k ≥ 1 A2k = 0 ∀ k ≥ 1

(4.11)

Equating coefficient of z 4 in (4.10 ) gives a0 a1 a2 a3 a4 + + + + =0 5! 4! 3! 2! 1! ha i a1 a2 a3 1 1 1 −1 0 a4 = − + + + =− + − = 5! 4! 3! 2! 120 48 72 720 Equating coefficient of z 6 in (4.10 ) a0 a1 a2 a3 a4 a5 a6 + + + + + + =0 7! 6! 5! 4! 3! 2! 1! From (4.11 )⇒ a3 = a5 = 0 a4 a2 a1 a0 − − − 3! 5! 6! 7! h 1 1 1 i 1 1 − + − = 3! 720 1.2.4.56 2.4.5.6 4.5.6.7 h i 1 1 1 = = 3! 6.4.5.6.7 6.7!

a6 = −

put A0 =

−1 ; A1 2

=

1 ; A3 12

f (z) =

=

−1 720

∞ X

=

−1 ; A5 6!

=

1 6.7!

and A2k = 0 for k ≥ 1 we get

An z n = A1 z −1 + A0 z 0 + A1 z + · · ·

n=−∞ ∞

=

4.1

1 1 X − + A2k−1 z 2K−1 z 2 k=1

Partial Fractions and Factorisation

Theorem: 36 (Mittag-Leffer Theorem). Let (bn ) be a sequence of distinct complex numbers such that limn→∞ bn = ∞ and pn (z) = an1 z + an2 z 2 + · · · + ankn z kn where 134

n = 1, 2, 3, · · · be arbitrary polynomials of degree atleast 1 and having no constant term. Then there are functions which are meromorphic in whole with poles at the points bn and the corresponding principal parts pn



1  an1 an2 an kn = + + ··· + 2 z − bn z − bn (z − bn ) (z − bn )kn

Moreover the most general moromorphic function can be written as f (z) =

∞ h  X pn n=1

i 1  − Qn (z) + g(z) z − bn

(4.12)

Where Qn (z) are suitably chosen fixed polynomials and g(z) is an entire function. Proof. Since the function pn



1 z−bn



is analytic for |z| < |bn |.

We can expand it by Taylor series about origin: pn



∞

1  X dns z s = z − bn s=0

Let us choose Qn (z) as a partial sum of pn



1 z−bn



(4.13)

and ending with term rn . There-

fore, Qn (z) =

rn X

dns z s

(4.14)

s=0

Pn − Qn can be calculated as follows: For this let us take the remainder term from Taylor series Z (z − z0 )n f (r)dr Pn − Qn = Rn = n 2πi c (r − z0 ) (r − z) n Z |z − z0 | |f (r)||dr| |Pn − Qn | = |Rn | ≤ n 2π c |r − z0 | |r − z| Put |z − z0 | = r,|f (z) ≤ M ,|r − z0 | = ρ and |r − z| ≥ |r − z0 | − |z − z0 | ≥ ρ − r Z rn M rn M (2πρ) |Pn − Qn | < |dr| = 2π c en .(ρ − r) 2π ρn .ρ − r  r n M ρ |Pn − Qn | ≤ ρ (ρ − r)

135

If we take ρ =

1 |b |; r 2 n

=

1 |b | 4 n

 and M = Mn = max 1

|Pn − Qn | ≤

|b | 4 n 1 |b | 2 n

n



1 pn z−b |z| n



=

1 |b | 2 n

Mn ( 21 |bn |) 1 |b | − 41 |bn | 2 n

2Mn 1 Mn ( 1 |bn |) = n = ( )n 1 2 2 2 |b | 4 n As n → ∞ ⇒ |Pn − Qn | → 0 (4.12 ) can be made convergent by choosing rn larger.

Now consider an arbitrary closed disc:|z| ≤ R. Since limn→∞ bn = ∞. This disc will only contain only a finite on of terms of (bn ). The series ∞ X n=1

 Pn

1 z − bn

 − Qn (z)

Has only a finite no of terms for which it becomes infinite in |z| ≤ R. Also (4.15 ) is true in |z| ≤ R. If the terms |bn | ≤ R are removed, if follow from weirestrass M-test the remaining series converges absolutely and uniformly in |z| ≤ R. Since R is arbitrary, the series (4.15 ) converges for all z 6= bn for n = 1,2, · · ·  and this represents a meromorphic function say h(z) if a particular term pn

1 z−bn

−

Qn (z) is removed from(4.15 ) , the resulting series converges to a function which is analytic and non-zero in some neighborhood of bn . Restoration of these terms show that f (z) has pole at bn with corresponding 1 principal part pn z−b clearly the only singularities of h(z) in the plane are z = n bn , n = 1, 2, l... If f (z) has same poles and the principal part as h(z) then the function g(z) is defined as g(z) = f (z) − h(z) has only removable singularities in complex plane and can be made analytic in the whole complex plane by suitably defined it at these singularities. We obtain f (z) = h(z) + g(z) where h(z) is defined by (4.15 ) and g(z) is analytic in the whole plane. (i.e) g(z) is an entire function. 136

Example ∞ P 2 1 Use M-L Therom to show that sinπ2 πz = deduce π cot πz = z1 + (z−n)2 n=−∞  ∞  ∞ P P 1 z 1 1 + n = z +2 where the prime to the summation sign indicates z−n z 2 −n2

n=−∞

n=1

that n takes all values except 0. Solution: Let f (z) =

π2 sin2 πz

The poles of f (z) are zero of sin2 πz ∴ sin2 πz = 0, ⇒ sinπz = 0, πz = ±nπ, z = ±n f (z) is meromorphic in the plane with poles of order a at each integer n. By direct continuation of Laurent’s expansion about z = 0, it is easy to see that the principal part of f (z) at the origin is

1 . z2

since sin2 πz = sin2 π(z − n). The principal part of f (z) at z = n is

1 . (z−n)2

Then

the series h(z) =

∞ X

1 (z − n)2 n=∞

(4.15)

converges for every z 6= n, n = 1, 2, · · · as can be seen by comparing it with familiar ∞ P 1 . It is uniformly convergent on any compact set after removal convergent series n2 1

of terms which becomes infinite of the set. ∴ As in the proof of M-L theorem, we can write ∞ X π2 1 − = g(z) 2 sin πz n=−∞ (z − n)2

(4.16)

Where g(z) is an entire function. We claim that g(z) ≡ 0 .To show this we observe f (z) and (4.15 ) are both periodic with period 1. Consequently, g(z) is also periodic

137

with p(z) = 1. Now | sin πz|2 = | sin π(x + iy)|2 = | sin πx cos(iπy) + cos πx sin(iπy)|2 = | sin πx cos hπy + i cos πx sin hπy|2 = sin2 πx cosh2 πy + cos2 πx sin h2 πy = cos h2 πy(1 − cos2 πx) + cos2 πx(cos h2 πy − 1) = cos h2 πy − cos2 πx

since cos h2 πy → ∞ as |y| → ∞. We see that

π2 sin2 πz

tends uniformly to zero as |y| → ∞. Also it can be shown that,

(4.15 ) tends uniformly to zero as |y| → ∞. Infacts, the convergence is uniform for |y| ≥ 1 and the limit for |y| → ∞ can be found by taking the limit of each term. ∴ We conclude g(z) = f (z) − h(z) tents uniformly to zero as |y| → ∞.(i.e) g(z) is bounded in whole plane. Hence by Liouvile’s theorem. g(z) must reduce to constant and lim|y|→∞ g(x + iy) = 0 we infer that g(z) = 0 Therefore, We obtain ∞ X π2 1 f (z) = h(z) ⇒ = 2 sin πz n=−∞ (z − n)2

*Deduction proof: we write (4.17 ) as, ∞ ∞ X π 1 1 X 1 = = 2 2 2 z n=−∞ (z − n)2 sin πz n=−∞ (z − n) n6=0

∞ X π2 1 1 − 2 = 2 (z − n)2 sin πz z n=−∞ n6=0

138

(4.17)

Using the expansion, 2 (πz)3 (πz)5 sin πz = πz − + + ··· 3! 5! i2 h (πz)2 (πz)4 + + ··· = π2z2 1 − 3! 5! ∞ 2 X π 1 1 ∴ − 2 = 2 (z − n)2 sin πz z n=−∞ 

2

n6=0

" 1  z2

#

1 1−

(πz)2 3!

+

(πz)4

+ ···

5!

2

1 − 2 = z

−2 (πz)2 (πz)4 1 1 − + + · · · − z2 3! 5!  −2 1 (πz)2 (πz)4 1 + 2 + + 3() + · · · − z2 3! 5!

1 = z2 1 = z2

1 1 + Higher power of z − − 2 = 2 z z function

π2 sin2 πz

∞ X

1 (z − n)2 n=−∞ ∞ X

1 (z − n)2 n=−∞ ∞ X

1 (z − n)2 n=−∞ ∞ X

1 (z − n)2 n=−∞

has a removable singularity a ∈ 0,as can be seen by direct calculation

the region D = (C − Z) ∪ {0} It is the derivative of f (z) =

1 z

− π cot(πz) ∴ f (0) = 0

Integrating,(4.17 ) Z

π2 dz = π 2 sin2 πz

Z

cosec2 πz dz

Z

d (− cot πz) dz π 2 (−π cot πz) = = −π cot πz π

= π

2

∞ X 1 1 ∴ −πcotπz + = − z z−n n=−∞

139

(4.18)

Since RHS is divergent, it can be made convergent by subtracting partial sum of the terms by n1 . Where n 6= 0   ∞  X X 1 1 1 1 1 1 − = − + =− + − z−n n z−n n z−n n n=∞ n=∞ n6=0  X 1 1 1 −πcotπz + = − + z z − n n n6=0   1 X 1 1 πcotπz = + + z n6=0 z − n n  ∞  1 X 1 1 = + + z n=1 z + n z − n ∞ X

Put n = −n in the 1st term and n = z − n in the 2nd , we get  ∞  1 1 1 X + πcotπz = + • z n=1 z − n n

Infinite products: If an infinite number of complex vectors are multiplied in given order according to some definite law, Then the product so formed is an infinite product. The product a1 , a2 , · · · an of finite number of complex vectors is denoted by and product of finite no of factors a1 , a2 , · · · is denoted by

∞ Q

n Q

ar

r=1

ar . For our conve-

r=1

nience, We take the factors of the form 1 + ar Convergence and divergence of infinite products Let an be a C number for all positive integral values of n and suppose pn = n Q

(1 + ar ) = (1 + a1 )(1 + a2 · · · (1 + an )). If Pn → a finite nonzero limit p as

r=1

n → ∞, we say infinite product ∞ Q

∞ Q

(1 + ar ) converges to the limit p and we write

r=1

(1 + an ) = p. If pn is finite nonzero limit, then

Q

(1 + an ) is said to be divergent Q in this case pn either tends to zero or infinity. When pn → 0 we say (1 + an )

r=1

diverges to 0. Q∞ Remark: n=1 (1 + an ) is said to converge iff atmost a finite no of factors are 140

zero and if partial products formed by non-vanishing factors tends to a finite limit different from zero. Q∞ n=1 (1+an ) is convergent, then pn and pn−1 must tend to same limit as n → ∞. pn = (1 + an )(1 + a2 ) · · · (1 + an ) pn−1 = (1 + a1 )(1 + a2 ) · · · (1 + an−1 ) pn = (1 + an ) → 1 as an → 0 pn−1 Q Hence necessary condition for convergence of (1 + an ) is that an → 0. It is not sufficient for example if an = n1 , Then pn n+1 1 → 1 as n → ∞ = (1 + an ) = 1 + = pn−1 n n But the product

∞ Q 1

(1 + n1 ) since pn = (1 + 1)(1 + 12 ) · · · (1 + n1 )

3 4 n+1 pn = 2. . · · · = n + 1 → ∞ as n → ∞ 2 3 n General principle of convergence Theorem: 37. The necessary and sufficient condition for convergence of an infinite product is that corresponding to any positive  however small, ∃ an integer m such that for all n ≥ m we have p + p n − 1 <  (whatever p may be) pn Proof. Necessary part: Let Pn converge to a finite non-zero limit P . Then a K > 0 can be found such that, |pn | > k∀n Now a number m can be chosen that pm+p − pm < k ∀integral values of p p k m+p − 1 < pm |pm | 141

(4.19)

from(4.19 )

Sufficient part: Assume pn+p pn

p m+p − 1 <  pm − 1 <  for n ≥ m, ∀p

pn+p −1< pn pn+p 1− < − 1 < 1 +  (Also true for m) pn − <

Since  can be chosen small that |pm+p | > (1 − )|pm |. Hence |pn | > some fixed positive no. When n > m and therefore p 9 0, also (1 − )pm < pm+p < (1 + )pm But pm is the product of finite no.of factors therefore and hence pm+p lies between two finite value (1 − e)pm and (1 + )pm ⇒ pn → finite limit as n → ∞ Q ∴ is convergent. Example: Show that (1 −

1 )(1 22

−

1 )··· 32

Converges to

1 2

Solution: Here 1 n2 2 n −1 (n + 1)(n − 1) = 2 n n2 2 3 8 n −1 (n + 1)(n − 1) . 2 ··· = 2 2 2 3 n n2 2 1.3 2.4 n −1 (n + 1)(n − 1) . 2 ··· = 2 2 2 3 n n2 1 1 1 (1 + ) Here pn−1 → as n → ∞ 2 n 2

a−n−1 = 1− = ∴ pn−1 = = = Also pn =

n(n + 2) 1 pn−1 → pn−1 → as n → ∞ 2 n 2 142

Theorem: 38. A necessary and sufficient condition for the convergence of ∞ Y

, (an 6= −1 for n = 1, 2, 3 · · · )

(4.20)

n=0

is the convergence of series

∞ P

log(1 + an ) where each logarithm has its principal

n=1

value.[or] The infinite product the series

∞ P

∞ Q

(1 + an ) with 1 + an 6= 0 converges simultaneously with

n=1

log(1 + an ) whose terms represent the values of principal branch of

n=1

logarithm. Proof. Necessary part: Let (4.20 ) converge so that limn→∞pn =p6=0,∞ exist where n Y (1 + ar ) pn =

(4.21)

r=1

Let us write sn =

n P

log(1 + ar ). Now (4.21 ) implies that

r=1

log pn =

n Y (1 + ar ) r=1

log pn + hn 2πi = sn

(4.22)

Where hn well determined integer. Since the principal value of logarithm is not necessarily same principal values of logarithm of its factors ⇒ hn is not necessarily 0. But hn is constant for sufficiently large n. Let αn , βn denote principle values of arg lof (1 + an ) and arg of pn respectively . Then we obtain βn + hn 2π = α1 + α2 + · · · + αn

(4.23)

βn+1 + hn+1 2π = α1 + α2 + · · · + αn+1

(4.24)

also

143

(4.24 )-(4.23 ), βn+1 − βn + 2π(hn+1 − hn ) = αn+1 ⇒ 2π(hn+1 − hn ) = αn+1 − βn+1 + βn → 0 (since αn+1 → 0 and βn+1 → βn as n → ∞) ⇒ hn+1 − hn → 0 as n → ∞ But hn is an integer ⇒ hn+1 = hn for sufficiently large n. If follows that hn is constantly equal to a certain integer h. Since pn → p ,We conclude sn → lim s = log P + h2πi Sufficient part: Assume that the series ∞ P log(1 + an ) converges to values limn→∞ sn = s n=1

Where sn =

n P

log(1 + an ). Consider

n=1

pn = (1 + a1 )(1 + a2 ) · · · (1 + an ) log pn = log(1 + a1 ) + · · · + log(1 + an )  X n n X log(1 + ar ) = exp(sn ) log(1 + ar ) ⇒ pn = exp = r=1

r=1

Since exp is a continues function then as sn → s ∴ pn → exp(s) 6= 0 Hence proved. Theorem: 39. A necessary and sufficient condition for absolute convergence of ∞ ∞ P Q (1 + an ) is the convergence of series |an |. 1

1

Proof. Necessary part: Let

P

|log(1 + an )| to be convergent so that

log(1 + an ) → 0 asn → ∞ log(1 + an ) → 1 asn → 0 144

⇒ an → 0 as n → ∞ Now we can find m such that n ≥ m and we have |an | < 21 , then a2 a2 log(1 + an ) = an − n + n + · · · 3  2  an a2n a3n = an 1 − + − + ··· 2 3 4 an a2n a3n log(1 + an ) −1 = − + − + ··· ⇒ a−N 2 3 4 Taking modulus, log(1 + an ) 1 ≤ |an | + 1 |an |2 + 1 |an |3 + · · · ] 2 an 3 4 since |an | < 21 log(1 + an ) 1 1 1 1 1 1 1 1 ⇒ − 1 < 2 + . 2 + . 3 < 2 + 3 + 4 + · · · an 2 3 2 4 2 2 2 2    −1 log(1 + an ) 1 1 1 1 1 − 1 < 2 1 + + 2 + · · · < 2 1 − an 2 2 2 2 2 1 2 < 2 = 2 2 1 log(1 + an ) 1 −1< < ⇒− 2 an 2 ⇒

1 log(1 + an ) 3 < −1< 2 an 2

from (4.25 ) we get 1 2 |an | ⇒ 2 ⇒ |an | X |an | Since

P

<

log(1 + an ) an

< | log(1 + an )| < 2| log(1 + an )| X < 2 |log(1 + an )|

| log(1 + an )| is convergent, by comparison test P ∴ |an | is convergent. 145

(4.25)

Sufficient part: P Assume |an | converges, then an → 0 as n → ∞ from( 4.25) log(1) + an 3 < an 2 2| log(1 + an )| < 3|an | X X 2 | log(1 + an )| < 3 |an | Since

P

|an | Converges.⇒

P

| log(1 + an )| converges.

Canonical products: A function which is analytic in whole plane is said to be entire integral functions. The simplest entire functions. Which are not polynomials are ez , sin z, cos z Result: If g(z) is a entire function, then f (z) = eg(z) is also entire function and not equal to zero. Conversely, if f (z) is any function which is never zero. Then f (z) is a from eg(z) . Converse part: Taking f (z) is an analytic function in the whole plane. Then, in entire C plane. ∴

f 0 (z) f (z)

is derivative of entire function f (z) (say)

(i.e.) f 0 (z) f (z)

=

d F (z) dz

146

f 0 (z) f (z)

6= 0 and analytic

Integrating with respect to z from z0 to z, Z z 0 Z z f (z) d dz = F (z)dz z0 f (z) z0 dz (F (z))zz0 = [log f (z)]zz0 F (z) − F (z0 ) = logf (z) − log(z0 ) log f (z) = F (z) − F (z0 ) + log f (z0 ) f (z) = exp(F (z) − F (z0 ) + log f (z0 )) Since, F (z0 ) and logf (z0 ) are constant and taking F (z) − F (z0 ) + logf (z0 ) = g(z) which is an entire function ⇒ f (z) = eg(z) Theorem: 40 (Weirestrass Theorem for Canonical Products). There exist an entire function with arbitrarily prescribed zeros an provided that in the case of infinitely many zeros an → ∞. Every entire func. with these and no other zeros can be written in the form, f (z) = eg(z) .z m .

∞ Q

(1 −

n=1

1 z 2 1 z mn z z )e an + 2 ( an ) +···+ mn ( an ) an

where product is taken for all an 6= 0, mn = certain integer and g(z) is an entire function. Proof. Let us first determine the form of most general entire function with finite no. of zeros. Assume that f (z) has m zeros at the origin and let the other zeros be denoted by a1 , a2 , · · · , aN multiple zeros repeated as many times as their orders. Then f (z) = z m (1 −

z )(1 a1

−

z ) · · · (1 a2

−

z )h(z) an

where h(z): entire func. with no zeros. By using previous result: h(z) = eg(z) where g(z): entire function N Q ⇒ f (z) = z m (1 − azn )eg(z) n=1

In the case of infinitely many zeros a1 , a2 , · · · , an , · · · with an 6= 0 and lim an = ∞ n→∞

and zero of order m at the origin of order n, we obtain similar representation by

147

means infinite products. The obvious generalization would be m

f (z) = z .

∞ Y

(1 −

n=1

z g(z) )e an

(4.26)

(4.26 ) holds if infinite product converges uniformly on every compact set. If this is so, then the product represents an entire func with zeros at the same points except for the origin and with same multiplicity as f (z). We conclude that the quotient can be written as z m eg(z) . ∴ product in (4.26 ) converges absolutely iff

∞ P n=1

1 |an |

converges and in this case the

converges is uniform in every closed disc |z| ≤ R. Under these circumstances we obtain (4.26 ). If (4.26 ) does not converge, we can construct the entire function corresponding to it by introducing convergence producing factors. We shall prove the existence of polynomial pn (z)  ∞  Y 1 − z pn (z) e a n n=1

(4.27)

Converges toan entire function. (4.27 ) converges iff, the series with general term  rn (z) = log 1−z + pn (z) converges where the branch or logarithm is chosen so an that the imaginary part of rn (z) lies between −π and +π. Let R > 0 be given and consider the terms with |an | > R in the disc |z| ≤ R, The principal branch of log( 1−z ) can be developed in a Taylor series an    2  3  mn  mn +1 z z 1 z 1 z 1 z 1 z log 1 − =− − − − ··· − − an an 2 an 3 an mn an mn + 1 an We chose  2  mn z 1 z 1 z pn (z) = + + ··· + an 2 an m n an

148

Then rn (z) = rn (z) = |rn (z)| ≤ ≤ = Since

mn +1 mn +2

 mn +1 1 z −pn (z) − − · · · + pn (z) m n + 1 an  mn +1  mn +2 1 z 1 z − − − ··· mn + 1 an m n + 2 an m +2 m +1 1 z n 1 z n − + − ··· an an mn + 2 an  mn +1  mn +2 1 z 1 R + + ··· mn + 1 |an | mn + 2 |an | m +1  2  1 R n mn + 1 R mn + 1 R 1+ + + ··· m n + 1 an mn + 2 an mn + 3 an

< 1 since mn+1 < mn+2 m +1  2  R R 1 R n 1 + + + ··· |rn (z)| ≤ mn + 1 an an an m +1  −1 R 1 R n |rn (z)| ≤ 1 − an mn + 1 an

(4.28)

Suppose the series ∞ X n=1

m +1 1 R n m n + 1 an

(4.29)

converges it follows from (4.28 ) that rn (z) → 0 as n → ∞ and consequently the imaginary part of rn (z) lies between −π and π as soon as n becomes sufficiently P large. Moreover, by Weirstrass M -test, rn (z) converges absolutely and uniformly on |z| ≤ R. Hence (4.27) represent analytic function in |z| < R. It remains to show that the series ∞ X n=1

 mn +1 1 R mn + 1 |an |

can be made convergent ∀R. By choosing mn = n and

R |an |

< 12 , |an | → ∞ can be

made to hold ∀ sufficiently large n so that,  n+1  n+1  n+1 1 R 1 1 1 < < n + 1 |an | n+1 2 2       ∞ ∞ n+1 n+1 X 1 R X 1 1 a ⇒ < → ( which is converges of 1−r = 12 ) n |a | 2 2 n n=1 n=1 149

and convergence of (4.29 ) is ensured because of convergence of geometric series  n+1 ∞ P 1 . Since R is arbitrary, (4.29 ). 2 n=1

Converges in the whole plane. If in addition to the points an . z = 0 is also a zero we have to introduce the factor z m in (4.27 ) and then the infinite product  ∞  Y z m 1− epn (z) z an n=1 defines an entire function. Whose only zeros are at the points 0, a1 , a2 , · · · finally if f (z) is an arbitrary entire function with prescribed zeros, then Zm

∞ Q n=1

f (Z)  1− az

n

epn (z)

represents entire function, without zeros and must be of the form eg(z) where g(z) entire function f (z)   = eg(z) z pn (z) z m Π∞ n=1 1 − an e   z m g(z) ∞ ⇒ f (z) = z e Πn=1 1 − epn (z) an   1 z 2 1 z mn z z m g(z) ∞ e an + 2 ( an ) +...+ mn ( an ) f (z) = z e Πn=1 1 − an

Corollary 3. Every function which is meromorphic in the whole plane is the quotient of 2 entire functions Proof Let F (z) be meromorphic functions in the whole plane. By the above theorem, we can construct an entire function g(z) having its zero at poles of F (z), then product F (z).g(z) defines an entire function F (z).g(z) = f (z) f (z) F (z) = g(z) Genus of canonical product We have m g(z)

f (z) = z e

Π∞ n=1

  z 1 z 2 1 z mn z 1− e an + 2 ( an ) +...+ mn ( an ) an 150

Now the product  ∞  Y z 1 z 2 1 z mn z 1− e an + 2 ( an ) +...+ mn ( an ) an n=1 represents an entire function. We shall choose all mn = each other, then the product becomes  ∞  Y z 1 z 2 1 z h z e an + 2 ( an ) +...+ h ( an ) 1− an n=1 This product converges kφ and represent an entire function if  h+1 ∞ X 1 R h + 1 |an | n=1 converges for all R provided that

∞ P



n=1

R |an |

h+1 < ∞. Assume that h is small integer

for which the series converges, then the expression   1 z 2 1 z n+1 z z ∞ Πn=1 1 − e an + 2 ( an ) +...+ n+1 ( an ) an is the canonical product associated with sequence an and h is called genus of canonical product. Result: Represent sin πz in the form of canonical product Solution: The zeros of sin πz are z = ±n where n = 1, 2, 3, . . . Here the genus of canonical P∞ 1 P 1 product is 1. Since the series ∞ n=1 n2 converges so that n = 1 n=1 n diverges and P 1 P 1 is the least integer such that = converges (n 6= 0) |±n|h+1 |n|h+1 Also z = 0 is a simple zero . Hence we obtain g(z)

sin πz = ze

 ∞  Y z z 1− en n n=−∞

(4.30)

where the prime to the product sign is used to indicate n takes all integral values except 0. We now determine g(z); 151

Taking log and diff (4.30 ) we obtain

∞  X

log(sin πz) = log z + g(z) +

n=−∞

π cot πz =

z z log(1 − ) + n n



∞ X

∞ X 1 1 1 −1 + g 0 (z) + ) + ( z z (1 − n ) n n n=−∞ n=−∞

π cot πz =

∞ ∞ X X 1 1 1 + g 0 (z) + + z z − n n=−∞ n n=−∞

(4.31)

note that the term by term differentiation is justified by uniform converges on any compact set which does not contain z = n we know that

 ∞  X 1 1 1 π cot πz = + + z n=−∞ z − n n substituting in (4.31 ) g 0 (z) = 0 ⇒ g(z) =constant from(4.30 ) we have ∞ Y sin πz z z g(z) =e (1 − )e n z n n=−∞ ∞ Y sin πz z z g(z) π =e (1 − )e n z n n=−∞ n6=0

we have sin πz =π z→0 z ∞ Y z z lim (1 − )e n = 0 z→0 n −∞ lim π

⇒ π = eg(z) we thus obtain

 ∞  Y z z sin πz = πz 1− en n −∞ n6=0

152

(4.32)

in this representation the factors corresponding to n and −n can be bracketed together so that

   ∞  Y z z z −z sin πz = πz 1− en 1 + en n n n=1  ∞  Y z2 = πz 1− 2 n n=1 It is evident from (4.32 ) that sin πz is an entire function of genus 1.

153

Chapter 5

Theorem: 41 (Riemann mapping theorem). Given any simply connected region Ω which is not the whole plane and a point z0 ∈ Ω, there exist a unique analytic function f (z) in Ω, normalised by f (z0 ) = 0, f 0 (z0 ) > 0 such that f (z) defines a 1-1 mapping of Ω onto the disc |z|. Proof. Suppose there exists two analytic functions f1 and f2 such that f1 (z0 ) = f2 (z0 ) = 0 and f10 (z0 ) > 0 and f20 (z0 ) > 0 defines 1 − 1 mapping of Ω onto the disc |ω| < 1. f1−1 and f2−1 exist (since f1 and f2 are 1 − 1) it maps |ω| < 1 onto the simply connected region Ω. ∴ (f1 f2 )−1 defines a mapping of |ω| < 1 onto itself. Such a mapping is given by a linear transformation. The general form is given by S(ω) =

eiα (ω − ω ¯0) ω 1−ω ¯0

From this, we have the condition S(0) = 0, and S 0 (0) > 0 (5.1) ⇒ S(0) = eiα (−¯ ω0 ) = 0 ⇒ω ¯0 = 0 Sub in (5.1 ), S(ω) = eiα ω S 0 (ω) = eiα ⇒ S 0 (0) = eiα > 0 154

(5.1)

If e(iα) = 1 ⇒ S(ω) = ω

∴ S defines an identity mapping. ∴ (f1 f2 )−1 (ω) = S(ω) = ω ⇒ (f1 f2 )−1 = Identity ∴ f1 = f2 .

Hence the mapping is unique. (Univalent function : An analytic function g(z) is said to be univalent if g(z1 ) = g(z2 ) ⇒ z1 = z2 ) Existence: Let J be the family of all functions g with following properties. (i) g is analytic and univalent in Ω. (ii) |g(z)| ≤ 1 on Ω. (iii) g(z0 ) = 0 and g 0 (z0 ) > 0 in Ω. We have to show that f is a member in J such that f 0 (z0 ) is maximum. (i.e) To prove : (a) J 6= φ. (b) there exist a function f with maximal derivative. (c) f has required properties. To prove (a): Since Ω is simply connected we can define single valued branch of function h(z) = √

z − a in Ω. This function does not have opposite value and does not have values

twice. ⇒ h(z) is a univalent function. Since Ω is open, by open mapping theorem, h(z) is open and it covers |ω − h(z0 )| < ρ. Also we know, 155

|ω − h(z0 )| < ρ cannot have intersection with |ω + h(z0 )| < ρ (i.e)|h(z) + h(z0 )| ≥ ρ contains z. Put z = z0 ⇒ |2h(z0 )| ≥ ρ

⇒ |h(z0 )| ≥

ρ 2

Now we can verify g0 (z) ∈ J : Consider g0 (z) =

ρ |h0 (z)| h(z0 ) h(z) − h(z0 ) . ∈J . 4 |h(z0 )|2 h0 (z0 ) h(z) + h(z0 )

g0 (z) is univalent since h(z) is univalent and g0 (z0 ) = 0. Now consider the estimate h(z) − h(z0 ) h(z) + h(z0 ) − 2h(z0 ) h(z) + h(z0 ) 2h(z ) 0 = = − h(z) + h(z0 ) h(z) + h(z0 ) h(z) + h(z0 ) h(z) + h(z0 ) 2h(z0 ) = 1 − h(z) + h(z0 )   1 2 ≤ |h0 | + |h(z) − h(z0 )| |h(z0 )|   2 2 ≤ + |h(z0 )| ρ ρ 4 = |h(z0 )| ρ ρ |h0 (z0 )| |h(z0 )| 4 |h(z0 )| ∴ |g0 (z)| ≤ 4 (|h(z0 )|)2 |h0 (z0 )| ρ g0 (z) ≤ 1 Now differentiate g0 (z) with respect to z,   d ρ |h0 (z0 )| h(z0 ) (h(z) + h(z0 ))h0 (z) − (h(z) − h(z0 ))h0 (z) (g0 (z)) = dz 4 |h(z0 )|2 h0 (z0 ) (h(z) + h(z0 ))2 ρ h0 (z0 ) h(z0 ) 2h(z0 )h0 (z) g00 (z) = 4 |h(z0 )|2 h0 (z0 ) h(z) + h(z0 )2 (i.e)g00 (z) ≥ 0 We have proved that g0 (z0 ) = 0 and g00 (z) =

ρ h0 (z0 ) h(z0 ) 2h(z0 )h0 (z) ≥ 0 ⇒ g0 ∈ J 4 |h(z0 )|2 h0 (z0 ) h(z) + h(z0 )2 156

∴ J 6= φ

To prove (b): To prove that there exists a function f in J with maximal derivative. Consider the set of derivative g00 (z0 ) with g0 ∈ J . Since g00 (z0 ) is positive, then the set must have a l.u.b say B. Therefore we can find sequence of function gn ∈ J such that g00 (z0 ) ≯ B Also |gn (z)| ≤ 1∀z ∈ ω. So the function in J must be uniformly bounded in every compact set. Therefore there exists a subsequence {gnk } of {gn } which converges to an analytic function f uniformly on every compact subset of Ω. Since each function of sequence is univalent, then the limit function is also univalent. For |f (z)| = | lim g(nk ) (z)| ≤ 1

(∵ |gn (z)| ≤ 1)

nk →∞

and f (z0 ) = limnk →∞ gnk (z0 ) = 0

(∵ gn (z0 ) = 0)

and f 0 (z0 ) = limnk →∞ gn0 k (z0 ) ≥ 0

(∵ gn0 (z0 ) > 0)

From this f 0 (z0 ) = B > 0 ⇒ f cannot be a constant. Let us choose some point z1 in Ω and consider the function g1 (z) = g(z) − g(z1 ) This function does not vanish identically so it is analytic. By Hurwitz Theorem, Every limit function is either zero or never zero. But the limit function f (z) − f (z1 ) which is not identically zero. ∴ f (z) − f (z1 ) 6= 0

⇒ f (z) 6= f (z1 )

⇒ z 6= z1 ∴ f is invariant. ∴ It has maximal derivative. To prove (c): To prove: f has required properties (i.e) To show f takes every value w with |w| < 1 Suppose that f (z) 6= w0 and |w0 | < 1 157

Since Ω is simply connected. We can define a single valued branch φ(z) =

q

f (z)−w0 1−w0 f (z)

Let φ(z1 ) = φ(z2 ) s ⇒

f (z1 ) − w0 = 1 − w0 f (z1 )

s

f (z2 ) − w0 1 − w0 f (z2 )

(f (z1 ) − ω0 )(1 − ω ¯ 0 f (z2 )) = (f (z2 ) − ω0 )(1 − ω ¯ 0 f (z1 )) (f (z1 ) − ω ¯ 0 f (z1 )f (z2 ) − ω0 − ω0 ω ¯ 0 f (z2 ) − f (z2 ) + ω ¯ 0 f (z2 )f (z1 ) + ω0 − ω ω ¯ 0 f (z1 ) = 0 f (z1 ) − f (z2 ) − ω0 ω ¯ 0 f (z2 ) + ω0 ω ¯ 0 f (z1 ) = 0 (1 − ω0 ω ¯ 0 )(f (z1 ) − f (z2 )) = 0 ⇒ f (z1 ) − f (z2 ) = 0 (Since|ω0 | < 1) f (z1 ) = f (z2 ) ⇒ z1 = z2 (Sincef is univalent) ∴ φ is Univalent. Next our aim is to show that φ(z)| ≤ 1 f (z) − ω0 2 = |f (z) − ω0 | |φ(z)| = 1−ω ¯ 0 f (z) |1 − ω ¯ 0 f (z)| Now |1 − ω ¯ 0 f (z)|2 − |f (z) − ω0 |2 = |1 − 2¯ ω0 f (z) + ω¯0 2 (f (z))2 | −|(f (z))2 + ω02 − 2ω0 f (z)| ≤ 1 − 2|¯ ω0 ||f (z)| + |¯ ω0 |2 |f (z)|2 − |f (z)|2 + |ω0 |2 −2|ω0 ||f (z)| (1 − |ω0 |2 f (z)) > |f (z) − ω0 |2 |f (z) − ω0 |2 1 > |1 − ω ¯ 0 f (z)|2 158

To normalise the function f consider a new function G(z)   |φ0 (z0 )| φ(z) − φ(z0 ) G(z) = ¯ 0 )φ(z) φ(z0 ) 1 − φ(z

(5.2)

G is univalent and also G(z0 ) = 0 Differentiate G(z) with respect to z ¯ 0 )φ(z))φ0 (z) − (φ(z) − φ(z0 ))(−φ(z ¯ 0 )φ0 (z)) |φ0 (z0 )| (1 − φ(z ¯ 0 )φ(z))2 φ(z0 ) (1 − φ(z ¯ 0) |φ0 (z0 )| φ0 (z) − φ(z0 )φ0 (z0 )φ(z = ¯ 0 )φ(z))2 φ(z0 ) (1 − φ(z ¯ 0 )) |φ0 (z0 )| φ0 (z0 )(1 − φ(z)φ(z = ¯ 0 )φ(z))2 φ(z0 ) (1 − φ(z

G0 (z) =

Then G0 (z) =

|φ0 (z0 )| φ0 (z0 ) . φ(z0 ) 1 − |φ(z0 )|2

(5.3)

From definition φ(z), s φ(z0 ) =

f (z0 ) − ω0 ) √ = −ω0 since f (z0 ) = 0 1 − ω¯0 f (z0 )

|φ(z0 )|2 = |ω0 | Differentiating φ(z) with respect to z,  − 12   1 f (z) − ω0 [(1 − ω ¯ 0 f (z))(f 0 (z)) − (f (z) − ω0 )(−¯ ω0 f 0 (z)) 0 φ (z) = 2 1−ω ¯ 0 f (z) (1 − ω ¯ 0 f (z))2  − 12  0  1 f (z) − ω0 f (z) − ω0 ω ¯ 0 f 0 (z) = 2 1−ω ¯ 0 f (z) (1 − ω ¯ 0 f (z))2 1 1 φ0 (z0 ) = (−ω0 )− 2 f 0 (z0 )(1 − |ω0 |2 ) 2

Since f 0 (z0 ) = B, B (1 − |ω0 |2 ) > B φ0 (z0 ) = √ 2 −ω0 ∴ G(z) becomes ∴ G0 (z) > B 159

which is a contradiction since B is the maximal derivative because f (z) is analytic and has the maximal derivative at z0 . This contradiction proves ∀ω0 with |ω0 | < 1, f (z) defines 1 − 1 mapping of simply connected region Ω into disc |ω| < 1. Boundary behaviour: The Jordan curve (simple, closed) divides the plane into exactly 2 regions, The one is bounded and other unbounded. The bounded region is Jordan region. we assume that f defines a conformal mapping of Ω onto another region Ω0 . If Ω and Ω0 are Jordan regions, then f can be extended to a topological mapping of closure of Ω onto closure of Ω0 . Definition: 29. Let Ω be a region and {zn } is a sequence in Ω. we say that {zn } tents to boundary of Ω (or) zn → ∂Ω if the points zn ultimately stay away from each point of Ω. ∴ Given z ∈ Ω ∃ > 0 and n0 ∈ N 3 |z − zn | ≥  for n 6= n0 Result: Let Ω be a region for sequence {zn } in ω, zn → ∂Ω iff given any compact K ⊂ Ω, ∃n0 ∈ N 3 these zn ∈ / k for n ≥ n0 . Proof. Sufficient part Since given z ∈ Ω we can find ρ > 0 3 the disc D(z, ρ) ⊂ Ω and we take k = D(z, ρ) and choose  < ρ. We get the proof. Necessary part: Let K be compact set ⊂ Ω. for each z ∈ K there exist z and nz such that |z−zn | ≥ z for n ≥ nn . The corresponding disc D(z, z ) = {r ∈ Ω|r − z| < z } is a open cover of K and by compactness, it has finite subcover N

k ⊂ ∪ D(ωi , ωi ) i=1

if n ≥ M = max nωk , then it is clear that i≤i≤N

zn ∈ / k for n ≥ M 160

Analytic Arcs: The complex function φ(t) defined in a < t < b is said to be real analytic if ∀ t0 , a < t0 < b φ(t) = φ(t0 ) + φ0 (t0 )(t − t0 ) +

1 (t − t0 )2 φ00 (t0 ) + · · · 2!

where it converges in some interval (t0 − ρ, t0 + ρ) for ρ > 0. Definition: 30. γ is said to be regular simple analytic arc if it given by a parametric equation φ(t), where φ(t) is real analytic, φ0 (t) 6= 0 and φ is 1-1. Definition: 31. γ is said to be free one-sided, regular, simple, analytic boundary arc of Ω if γ is regular, simple arc given by parametric equation φ(t) which is real analytic and can be extended to a simply connected region ∆ symmetric to (a, b) with property that φ(a) ∈ ω, whenever t lies in upper half of ∆ and φ(t) lies outside ω if t lies in lower half of ∆. Definition: 32. Let Ω be a region such that ∂Ω contains a line segment γ, with parametric equation γ(t) = (1 − t)ζ + t, t ∈ (0, 1). we say γ is free boundary arc if for every ζ ∈ γ and sufficiently small r > 0, we have B(ζ, r) ∩ ∂Ω is a actual diameter of B(ζ, r) if γ is free boundary arc for ζ ∈ γ, then ∃ r0 > 0, B(ζ, r) ∩ γ is diameter of B(ζ, r) for every r ∈ (0, r0 ). hence the 2 open half discs say H1 and H2 determined by diameter does not intersect ∂Ω, then we have 2 possible cases: • H1 ⊂ Ω and H2 ⊂ Ωc • H1 ⊂ Ω and H2 ⊂ Ω

Definition: 33. Let ζ ∈ γ and r > 0 be such that B(ζ, r) ∩ ∂Ω is the diameter of B(ζ, r) • if one of 2 half open disc determined by diameter of B(ζ, r) lies in Ω and the other half disc lies in Ωc , then ζ is a one-sided free boundary point of Ω. 161

• If both half open discs determined by diameter of B(ζ, r) lie in Ω, then ζ is a 2-sided boundary point of Ω. Definition: 34. Let γ be a free boundary arc in Ω, then • ζ is a one-side free boundary point of Ω for all ζ ∈ γ iff γ is a one- side free boundary arc. • ζ is 2-side boundary point of Ω for all r ∈ γ iff γ is 2-side free boundary arc. Conformal mapping of polygons: Let us assume Ω be a bounded, simply connected region whose boundary is a closed polygonal line without self intersection. Let the consecutive  intersections be z1 , z2 , · · · zn . The angle at zk is given by the  k−1 −zk between 0 and 2π, we denote it by αk π, the corresponding exvalue arg zzk−1 −zk terior angle βk π ⇒ αk π + βk π = π ⇒ αk + βk = 1 ∴ βk = 1 − αk and also β1 π + · · · + βn π = 2π n X βk = π ∴ k=1

The polygon is said to convex iff all βk ≥ 0. Theorem: 42 (Schwarz-christoffel’s formula). The function z = F (w) which maps |ω| < 1 conformally onto polygons with angles αk π(k = 1, 2, · · · n) are of the form Z ωY n F (w) = c (ω − ωk )βK dω + c0 where βk = 1 − αk 0

k=1

The ωk are points on the unit circle and c, c0 are constants Proof. Let Ω bounded, simply connected region whose boundary is a closed polygons with interior angles αk π(k = 1, 2, · · · n) We know that a mapping f (z) can be extended by continuity to any side of polygon and that each side is mapping in 1 − 1 onto arc of the unit circle. 162