Elementary Functions

x Q. If u ( x, y ) = 2 , find a 2 x +y harmonic conjugate v of u. Soln : Observe the following :

1 (i ) If f ( z ) = , then u = Re f ( z ). z (ii ) f ( z ) is analytic in a domain D = C - {(0, 0)}. y (iii) Im f(z) = v = − 2 . 2 x +y Conclude that v is a H.C. of u.

Chapter 3: Elementary Functions 1. 2. 3. 4. 5.

Exponential Functions Trigonometric Functions Hyperbolic Functions Logarithmic Functions Complex Exponents

1. Inverse Trigonometric Functions 2. Inverse Hyperbolic Functions

3

n

z z = 1 + z + + + ...... + + .... 2! 3! n! z laurin'series of e

=e

z

2

z1 + z2

,

z1 e

e

z2

z1 − z2 =e

z

(2) Let f(z) = e = e x

= e .e

x + iy

iy

x

= e (cosy + i sin y) ≡ u + iv x

x

⇒ u = e cos y, v = e sin y,

x

x

⇒ u x = e cos y, u y = −e sin y x

x

v x = e sin y, v y = e cos y ⇒ u x = v y , u y = −v x

Thus CR equations are satisfied and clearly u x , u y , v x , v y are continuous

⇒ f ( z ) is differentiable and f ′(z) = u x + i v x x

x

x

iy

= e cos y + i e sin y = e . e = e d z z ⇒ e =e dz

( )

z

z

x

(3) e = e . e

iy

iy

e = cos y + i sin y ⇒e

iy

z

2

2

= cos y + sin y = 1 x

x

x

∴ e = e = e as e > 0 ∀ x ∈ R z

⇒ e ≠ 0 for any complex number z.

iφ

We may write e = e . e = ρ e , z

x

iy

when ρ = e = e > 0 & φ = y x

( )

z

∴ arg e = y + 2nπ , n = 0, ± 1, ± 2........... z

(4)

cos 2π = 1 & sin 2π = 0

Hence e πi

2πi

= cos 2π + i sin 2π = 1

e = cos π + i sin π = −1 −π i e = cos( − π ) + i sin ( − π ) = −1

π i/2

e

e

−π i / 2

= cos π / 2 + i sin π / 2 = i = cos ( − π / 2) + i sin ( − π / 2) =−i

5. e

z + 2πi

z

= e .e

2πi

=e

z

z

⇒ e is periodic with imaginary period 2πi.

∴e

z ± 2 nπi

z

= e ∀ n = 0,1,2,3,............

x

(6) e > 0 ∀ x ∈ℜ z

But e may be negative if z ∈C z

Example : Find z such that e = −1

Solution : z

e = −1 x

iy

πi

⇒ e . e = 1.e x

⇒ e = 1, and y = π + 2nπ , n = 0, ± 1,. ± 2... ⇒ x = 0 & y = π + 2 nπ

Thus, if z = x + iy = (2n + 1)π i, n = 0, ± 1,± 2,... z

then e = − 1

Excercise : z

(7) e is not analytic anywhere.

Q. Find all values of z such that e

2z −1

= 1+ i

Solution : 2 z −1 e = 1+ i ⇒e

2 x −1

.e

2iy

= 2

π i 4 e

⇒e

2 x −1

= 2,

π 2 y = + 2nπ ; 4 n = 0,±1, ± 2,..

(

)

1 π ⇒ x = 1 + ln 2 , y = + nπ 2 8 ∴z = x + iy

(

)

1 π  = 1 + ln 2 + i + nπ , 2 8  n = 0, ± 1, ± 2,....

Trigonometric Functions (1) If x is real , then ix

−ix

e +e cos x = , 2 ix −ix e −e sin x = . 2i

If z is complex, we define iz

−iz

e +e cos z = , 2 iz −iz e −e sin z = − − − (1) 2i iz ⇒ e = cos z + i sin z ,

sin z cos z tan z = , cot z = , cos z sin z 1 1 sec z = , cos ec z = cos z sin z

2. Since ez is analytic ∀ z and linear

combination

analytic

functions

of is

two again

analytic, hence it follows that sin z and cos z are analytic functions.

3.Using (1) it is easy to prove : i) sin( −z ) = − sin z ii) cos(−z ) = cos z d ( sin z ) = cos z iii ) dz

iv )

d ( cos z ) = − sin z dz

v)

d 2 ( tan z ) = sec z dz

vi) sin( z1 ± z2 ) = sin z1 cos z2 ± cos z1 sin z2 vii) cos( z1 ± z2 ) = cos z1. cos z2 sin z1 sin z2

( 4)

iz

−iz

e +e cos z = , 2 iz −iz e −e sin z = . 2i

Put x = 0, then

i ( iy )

− i ( iy )

+e cos( iy ) = 2 −y y e +e = = cosh y 2 e

1 y −y sin ( iy ) = − (e − e ) 2i 1 y −y = i (e − e ) 2 = i sinh y

cos z = cos( x + iy ) = cos x cos iy − sin x. sin iy = cos x . cosh y − i sin x . sin hy

sin z = sin ( x + iy ) = sin x. cos iy + cos x. sin iy = sin x. cosh y + i cos x . sin hy

Hence (EXCERCISE) 2

2

2

sin z = sin x + sinh y 2

2

2

cos z = cos x + sin h y

Hints : (Use) 2

2

cos h x − sinh x = 1 2

2

cos x + sin x = 1

5.Analyticity of tan z & sec z : sin z 1 tan z = , sec z = cos z cos z ⇒ tan z & sec z are analytic everywhere except at the points where cos z = 0

cos z = 0 ⇒ cos ( x + iy ) = cos x cos hy − i sin x sin hy = 0 ⇒ cos x cos hy = 0, & sin x sinh y = 0

cosh y ≠ 0 y

−y

e +e 1 y 1  ( cosh y = = e + y  2 2 e  2y

= 0 ⇒ e = − 1 < 0)

π ∴ cos x ⇒ 0 ⇒ x = ( 2n + 1) , n = 0, ± 1, ± 2... 2

π But sin x ≠ 0 for x = ( 2n + 1) 2 ∴ sinh y = 0 ⇒ y = 0   e −e 2y = 0 ⇒ e = 1 ⇒ y = 0  sinh y = 2   y

−y

π ∴ z = x + iy = ( 2n + 1) 2 ∴ tan z & sec z are analytic every where except at π z = ( 2n + 1) , n = 0, ± 1 ± 2,..... 2

(6 Ex.) Analyticity of cot z & cosec z : cos z 1 cot z = & cos ec z = sin z sin z ⇒ cot z & cos ec z are analytic every where except at the points where

sin z = 0

sin z = sin ( x + iy ) = sin x cosh y + i cos x. sinh y = 0 ⇒ sin x. cosh y = 0 & cos x sinh y = 0 cosh y ≠ 0 ⇒ sin x = 0 ⇒ x = nπ , n = 0,± 1, ± 2,...

But for

x = nπ ,

cos x ≠ 0

∴sinh y = 0 ⇒ y = 0 ∴z = x + iy = nπ Thus cot z & cos ec z

are

analytic everywhere except at the points where z = nπ ,

n = 0, ±1, ± 2,...

Hyperbolic Functions : Definition : z

−z

e −e sinh z = , 2 z −z e +e cos hz = . 2

(1) e

z

&e

−z

are analytic everywhere

⇒ sin h z & cosh z are analytic everywhere.

d d e − e  (2) [ sin h z ] =   dz dz  2  z

z

−z

−z

e +e = = cos h z 2 d [ cos hz ] = sin hz Similarly dz

3. cos z = cos h( i z ) , z

−z

e +e cos hz = 2 iz −i z e +e ⇒ cos h( i z ) = = cos z 2

4. cos( i z ) = cosh z cos z = cos h( i z ) 2 ⇒ cos( i z ) = cos h i z = cos h( − z ) = cosh z

( )

5.

sin z = − i sin h ( i z )

6.

sin ( i z ) = − i sin h ( − z ) = i sin h z

7. sinh ( z1 + z2 ) = sin h z1. cos h z2 + cos h z1. sin h z2 8. cos h( z1 + z2 ) = cos h z1. cos h z2 + sin h z1. sin h z2

9. sin h( − z ) = − sin h z cos h ( − z ) = cos h z 2 2 cos h z − sin h z = 1

(10) sin h z = sin hx . cos y + i cos hx . sin y Soln :

sin h z = sin h( x + iy ) = sin hx cos h( i y )

+ cos h x. sin h( i y ) = sin hx. cos y + i cos hx . sin y

Excercise : 2

2

2

sin h z = sin h x + sin y

Similarly a) cosh z = cos hx cos y + i sin hx. sin y 2

2

2

b) cosh z = sin h x + cos y

(11) Analyticity of tan hz & sec h z : sinh z tan h z = , cos hz 1 sec hz = . cos hz

⇒ tanh z & sec hz are analytic everywhere except at the points where cos h z = 0.

Now cosh z = 0 ⇒ cos ( i z ) = cos( ix − y ) = 0 ⇒ cos ( i x ). cos ( y ) + sin ( ix ). sin ( y ) ⇒ cosh x . cos y + i sinh x. sin y = 0 ⇒ cosh x. cos y = 0, sinh x. sin y = 0.

cos h x ≠ 0 ⇒ cos y = 0

π ⇒ y = ( 2n + 1) , n = 0, ± 1, ± 2,... 2 π For y = ( 2n + 1) , sin y ≠ 0 2 ∴sin h x = 0 ⇒ x = 0

∴ z = x + iy iπ = ( 2n + 1) , 2 n = 0, ± 1, ± 2, ...

⇒ tan hz & sec hz are analytic everywhere except at iπ z = ( 2n + 1) , n = 0, ± 1, ± 2,.... 2

Exercise: coth z and cosech z are analytic everywhere except at z = nπi,

n = 0, ± 1, ± 2,.....

Q. Show that (i ) sin h ( Im z ) ≤ sin z ≤ cos h( Im z ) (ii ) sin h( Im z ) ≤ cos z ≤ cos( Im z )

Sol : (1) sin z = sin ( x + iy ) = sin x. cos( iy ) + cos x. sin ( iy ) = sin x. cosh y + i cos x. sin h y

2

2

2

⇒ sin z = sin x. cos h y 2

2

(

2

+ cos x. sin h y 2

= sin x 1 + sin h y

(

2

)

) 2

+ 1 − sin x . sin h y 2

2

= sin x + sin h y

2

2

2

2

⇒ sin h y ≤ sin z = sin x + sin h y 2

≤ 1 + sinh y 2

= cosh y ⇒ sin hy ≤ sin z ≤ cosh y

(ii ) cos z = cos( x + iy ) = cos x. cos( iy ) − sin x. sin ( iy ) = cos x. cosh y − i sin x. sin h y

2

2

2

⇒ cos z = cos x. cos h y 2

2

+ sin x. sin h y 2

(

2

= cos x 1 + sin h y

(

2

)

)

2

+ 1 − cos x . sin h y 2

2

= cos x + sin h y

2

2

2

2

⇒ sin h y ≤ cos z = cos x + sin h y 2

≤ 1 + sin h y 2

= cosh y ⇒ sin h y ≤ cos z ≤ cos hy

The Logarithmic Function : The natural logarithm of z = x + iy is denoted by log z, i.e. w = log z,

and log z is defined for z ≠ 0 by the relation w

e = z ..............(i ) w

i.e. if e = z , then we write w = log z

Let w = u + iv, z = x + iy = r cos Θ + i r sin Θ iΘ

= r e , where − π < Θ ≤ π , Θ = Arg z

Then (i ) ⇒ e

u + iv

⇒e . e = r e u

iv

=re

iΘ

⇒e = r = z, u

v = Θ + 2 nπ , n = 0, ± 1, ± 2,......

iΘ

⇒ u = ln r = ln z , v = Θ + 2nπ

∴ w = log z = u + i v = ln z + i( Θ + 2nπ )

Since Arg z = Θ, − π < Θ ≤ π and arg z = Θ + 2nπ , n is any integer ∴log z = ln z + i arg z ,

z≠0

When n = 0, then arg z = Arg z When n = 0, then the value of log z is called the principal value of log z and is denoted by Log z, i.e. Log z = ln z + i Arg z , z ≠ 0.

∴ log z = ln z + i arg z = ln z + i( Θ + 2nπ )

= ( ln z + iΘ ) + i 2nπ ⇒ log z = Log z + i 2nπ , n = 0, ± 1,± 2,...

Sec 31 : If z1 & z 2 be any two non − zero complex numbers, then (1) log( z1 z 2 ) = log z1 + log z 2  z1  (2) log   = log z1 − log z 2  z2 

But Log ( z1 z 2 ) ≠ Log z1 + Log z 2  z1  Log   ≠ Log z1 − Log z 2 z  2 n

Log z ≠ n Log z

Ex (1) Let z1 = −1, z 2 = −1 z1=-1+i0

z1

z1 z2

∴Log ( z1 ) = ln z1 + i Arg z1

⇒ Log ( − 1) = ln (1) + i Arg z1

= 0 + iπ ∴Log ( z1 ) + Log ( z 2 ) = 2π i

But z1 z2 = 1

⇒ Log ( z1 z2 ) = ln z1 z2 + i Arg ( z1 z2 ) = 0 + i. 0 = 0 Thus Log ( z1 z2 ) ≠ Log z1 + Log z2

Q.3(b) Log ( − 1 + i ) ≠ 2 Log ( − 1 + i ) 2

L.H.S. = Log ( − 1 + i )

[

2

2

= Log 1 + i − 2 i = Log ( − 2 i )

]

= ln − 2i + iArg ( − 2i )  π = ln 2 + i −   2 π = ln 2 − i 2

RHS = 2 Log ( −1 + i )

= 2[ln −1 + i + i Arg ( −1 + i ) ] 3π   = 2 ln 2 + i  4  

1 3π = 2[ ln 2 + i ] 2 4 3π = ln 2 + i 2 ∴LHS ≠ RHS

Q.4 Show that

( a ) log(i

2

) ≠ 2 log i,

log z = ln r + i θ ,

r = z > 0, 3π 11π <θ < 4 4

when

( ) 2

(b) log i = 2 log i, when log z = ln r + iθ , π 9π r = z > 0, < θ < 4 4

Soln (a) : LHS = log(i ) = log( −1) 2

= ln −1 + i θ , = 0 +π i = π i

NOTE : 3π / 4 < θ < 11π 4 ,

θ = arg( − 1) = π + 2nπ and hence n = 0.

We have log i = ln i + i arg i π  = ln 1 + i + 2nπ , where 2  n is an integer 1  = iπ  2n +  2 

3π 1 11π  < θ =  2n + π < 4 2 4  ⇒ n = 1 & hence 5π θ= 2

11 π/4

3 π /4

5π ∴ RHS = 2 log i = 2. i = 5π i 2 LHS =/ RHS

Soln(b) : π 9π < θ = ( 2n + 1)π < 4 4 ⇒n=0 ⇒ LHS = πi

9π/4 π/4

But when

π 1 9π  < θ =  2n + π < 4 2 4  π ⇒ n = 0 & hence θ = 2

π RHS = 2 log i = 2 i = πi 2 ∴LHS = RHS

( ) = 2 log i

i.e. log i

2

π 9π if <Θ< 4 4

Sec 30 : Derivatives of log z and Log z

Remark 1 : Since log z = ln z + i arg z = ln z + i ( Θ + 2nπ ) , n = 0, ± 1, ± 2,..... ⇒ log z is a multivalued function.

Remark 2 : Since Log z = ln z + i Θ , Θ = Arg z ⇒ Log z is a single - valued function.

Re mark 3 :

(

)

1 2 2 ln z = ln x + y 2 is continuous everywhere except at ( 0,0).

Remark 4 : Let α be any real number, and consider f ( z ) = log z = ln z + iθ = ln r + iθ ,

( r > 0,α < θ < α + 2π ) ⇒ u ( r ,θ ) = ln r , v( r ,θ ) = θ

y

α

x

Then log z is single - valued and continuous in the domain D = {z : z > 0, α < θ < α + 2π }

Remark 5: The function log z is NOT continuous on the line

θ =α

as arg z is NOT θ = α continuous on the line

.

For if z is a point on the ray θ=α then there are points arbitrary close to z at which the values of v are nearer to α, and also there are points such that the values of v are nearer to α+2π.

⇒ lim arg z does not exist. z →α

Remark 6 : (i) log z = ln r + i θ

is analytic

in domain D1 = {z : z = r > 0, α < θ = arg z < α + 2π } (ii ) Log z = ln r + i Θ is analytic in the domain D 2 = {z : z = r > 0, − π < Θ = Arg z < π }

As, u( r ,θ ) = ln r , v ( r ,θ ) = θ 1 ⇒ ur = , uθ = 0 r v r = 0, vθ = 1

⇒ CR - equations in polar form r u r = vθ , uθ = −r vr are satisfied and first - order partial derivatives are continuous.

d − iθ ⇒ f ′( z ) = ( log z ) = −e ( ur + i vr ) dz 1 1 = iθ = in D1 z re In particular, when α = −π d 1 ( Log z ) = in D2 . dz z

Remark : 7 Log z is analytic on the whole

complex plane except at ( 0,0 ) and on the ray θ = -π , i.e. on negative real axis.

i.e. singularties of Log z are given by Re z ≤ 0, Im z = 0.

Definition : A branch of a multiple - valued function f ( z ) defined on a set S is any single valued function F(z) that is analytic in some domain D ⊆ S such that for all z ∈ D, F( z ) is one of the values of f ( z ).

Ex. For each fixed α , log z = ln z + i θ ,

( z > 0,α < θ < α + 2π )

is a branch of log z = ln z + i arg z

Log z = ln z + i Θ ,

( z > 0, − π < Θ < π ) is called the principal branch.

Q.9(a) p. 94 Show that the function Log ( z − i ) is analytic everywhere except on the half line y = 1 ( x ≤ 0 ).

Solution : We have f ( z ) = Log ( z − i ) singularity of f ( z ) is given by

Re ( z − i ) ≤ 0 & Im( z − i ) = 0 ⇒ Re( x + i ( y − 1) ) ≤ 0 & Im( x + i ( y − 1) ) = 0

⇒ x ≤ 0 & y =1

y=1

Q 9 ( b ) Show that the function Log ( z + 4 ) f ( z) = 2 z +i is analytic everywhere except at the points ± (1 - i ) / 2 and on the portion x ≤ − 4 of the real axis.

Solution : Singularities of f ( z ) are given by Re( z + 4 ) ≤ 0, Im( z + 4 ) = 0 & 2

z +i = 0 2

⇒ x + 4 ≤ 0, y = 0 & z = −i

2

Now z = − i = e

⇒z =e ⇒z =e

 −π  + 2 nπ  i   2 

, n = 0, 1

 −π i + 2 nπ    2 2  −π  + nπ  i   4 

, n = 0,1

When n = 0, then z=

−π i 4 e

π π = cos − i sin 4 4 1 (1 − i ) = 2

When n = 1, then z=e

 π  π − i 4 

 3π   3π  = cos  + i sin    4   4  1 (1 − i ) =− 2

Hence singularities of f ( z ) are 1 (1 − i ) , x ≤ −4. ± 2

Sec 32 : Complex Exponents (1) Let z ≠ 0 be a complex no., and c is any complex no. c

Then z is defined as c

z =e

c log z

If log z is replaced by Log z, then c

z =e

c Logz

is called the principal value c

of z .

Q.2( a ) Show that i is real and find its principal value. i

i

Solution : i = e

i log i

log( i ) = ln i + i arg ( i ) 1 π   = 0 + i + 2nπ  =  2n + π i 2 2  

i

∴ i =e

1  −  2 n + π 2 

which is real i

Principal value of i is π − 2 e

(n = 0).

EX. ( b ) Find P.V. of i . -i

Solution : −i

i =e =e

−i log i

=e

1   2 n + π 2 

1  −i  2 n + πi 2 

, n = 0, ± 1, ± 2,.. −i

π 2

Principal value of i = e

(c) Write log(Log i) in terms of a + ib

π We have Log i = i (WHY ??) 2 π π  π  ⇒ log ( Log i ) = log i  = ln i + i arg i  2 2  2  π  = ln ( π 2 ) + i + 2nπ  2  Principal value of π Log ( Log i ) = ln ( π 2 ) + i 2

Q. Find the principal value of (1 - i )

( 1+i ) log ( 1−i ) =e

1+i

Solution : (1 − i ) Now log (1 - i ) = ln 1 − i + i arg (1 − i )  π  = ln 2 + i − + 2nπ   4  1+i

1

-i

-π / 4

∴ (1 - i ) =e

1+i

= e

log ( 1−i )

.e

= (1 − i ) . e

log ( 1−i ) + i log ( 1−i )

i log ( 1−i )

1  i ln 2 −  2 n − π 4 

i ln 2

1  − 2 n − π 4 

= (1 − i )

π i ln 2 4 e e

= (1 - i ) e .e Principal value of

(1 - i )

1+i