16 Laurent & Taylor's Expansion Recidue

Taylor’s Theorem: Suppose that a function f(z) is analytic throughout a disk z − z0 < R0 centered at z0 and with radius R0. Then f(z) has the power series representation

∞

f ( z ) = ∑ an ( z − z 0 ) , n= 0

n

( z − z0 < R0 )

where an =

f

( n)

( z0 )

n!

(n = 0,1, 2.....)

Maclaurin Series

Taylor Series about the point z 0 = 0 is called Maclaurin series, i. e. ∞

f ( z) = ∑

n =0

f

(n)

( 0) n z , ( z < R0 ) n!

Examples:

1. 2.

∞

n

z e = ∑ , ( z < ∞) n ! n= 0 z

∞

2 n +1

z sin z = ∑ (− 1) , ( 2 n + 1 )! n= 0 n

( z < ∞)

∞

2n

z 3. cos z = ∑ (− 1) , ( 2 n )! n= 0 n

( z < ∞) 4.

∞

2 n +1

z sinh z = ∑ , n = 0 ( 2n + 1)! ( z < ∞)

5.

∞

2n

z cosh z = ∑ , ( 2 n )! n= 0 ( z < ∞)

6.

∞

1 n =∑z , 1 − z n= 0

( z < 1)

∞

1 n n 7. = ∑ (− 1) z , 1 + z n= 0 ( z < 1)

Laurent’s Theorem: Suppose that a function f(z) is analytic throughout an annular domain R1 < z − z0 < R2 centered at z and 0 let C denote any positively oriented simple closed contour around z0 and lying in that domain.

Then, at each point in domain f(z) has the series representation ∞

f ( z ) = ∑ an ( z − z 0 ) n =0

bn +∑ n n =0 ( z − z0 )

(R < z − z 1

where

n

∞

0

< R2 )

1 f ( z ) dz an = ( n = 0 , 1 , 2 .....) ∫ n +1 2π i C ( z − z0 )

and 1 f ( z ) dz bn = ( n = 0 , 1 , 2 ,...) ∫ − n +1 2π i C ( z − z0 )

Example:

Find the Laurent series representation of z f ( z) = ( z − 1)( z − 3) when

(a) D1 : 0 < z < 1, (b) D 2 : 1 < z < 3, (c) D3 : 3 < z < ∞,

We have z f(z) = (z − 1)(z − 3) 1 3 =− + 2( z − 1) 2( z − 3)

(a) Consider the domain D1 : 0 < z < 1. Then f(z) is analytic in D1.

1 3 f(z) = − + 2( z −1) 2( z − 3) 1 3 = − 2(1 − z ) 2 × 3(1 − z ) 3 ∞

∞

1 1 z n = ∑ z − ∑  2 n =0 2 n =0  3 

n

∞

∞

1  z n 1 ⇒ f(z) = ∑ z − ∑   2 n= 0 2 n= 0  3  ∞

1  1 n = ∑ 1− n  z 2 n= 0  3 

n

(b) Consider the domain D 2 : 1 < z < 3. Then f(z) is analytic in D 2 .

1 3 f(z) = − + 2( z − 1) 2( z − 3) 1 3 =− − 1 z 2 z (1 − ) 2 × 3(1 − ) z 3 ∞

n

∞

1 1 1  z = − ∑  − ∑  2 z n =0  z  2 n =0  3 

n

∞

∞

1 1 1  z ⇒ f(z) = − ∑ n+1 − ∑   2 n= 0 z 2 n= 0  3 

n

(c) Consider the domain D 3 : 3 < z < ∞. Then f(z) is analytic in D 3 . Note that 1 3 < < 1. z z

1 3 f(z) = − + 2( z − 1) 2( z − 3) 1 3 =− + 1 3 2 z (1 − ) 2 × z (1 − ) z z ∞

n

∞

1 3 1  3 = − ∑  + ∑  2 z n =0  z  2 z n =0  z 

n

∞

n

∞

1 3  3  1 ⇒ f(z) = − ∑   + ∑   2 z n= 0  z  2 z n= 0  z  ∞

∞

n +1

1 1 1 3 = − ∑ n +1 + ∑ n +1 2 n= 0 z 2 n= 0 z ∞

(

)

1 n +1 1 = − ∑ 1− 3 . n +1 2 n= 0 z

n

Excercise:

Show that, when 0 < z − 1 < 2, the Laurent series representation of z f ( z) = ( z − 1)( z − 3) is

∞

n

( z − 1) 1 f ( z ) = − 3∑ n+ 1 − . 2 ( z − 1 ) 2 n= 0

RESIDUE

(1) Consider a function f(z) & 1 let z = . Then w 1 f ( z ) = f   = g ( w) w

(i ) f(z) is said to be analytic at infinity if g(w) is analytic at w = 0. (ii) f(z) is said to be singular at infinity if g(w) is singular at w = 0.

(2) Zero of an analytic function : Let f(z) is analytic in a domain D. If f(z 0 ) = 0 for some z = z 0 , then z = z 0 is called zero of f(z).

If f(z 0 ) = f ′( z0 ) = f ′′( z0 ) = .... = f f

(n)

( n −1)

( z0 ) = 0,

but

( z0 ) ≠ 0, then z = z0 is

called ZERO OF ORDER n of f ( z ).

i.e. z = z0 is called zero of order n of f(z) if n

f ( z ) = ( z − z0 ) g ( z ), where g ( z0 ) ≠ 0.

(3)Singular Point of a fn f(z) : (i) If a function f(z) fails to be analytic at a point z 0 , but it is analytic at some point in every nbd of z 0 , then z 0 is called Singular Point of f(z).

(ii) Isolated Singularity The point z 0 is called an isolated singularity of f(z) if (a) z 0 is a singular point of f(z) (b) f(z) is analytic in a deleted nbd N : 0 < z - z 0 <∈ .

(4) (i) Let z 0 is an isolated singularity of f(z) ⇒ ∃ R > 0 such that f(z) is analytic in 0 < z − z0 < R.

Hence f(z) has Laurent series expansion : ∞

∞

f(z) = ∑ an ( z − z0 ) + ∑ bn ( z − z0 ) , n= 0

n

n= 0

0 < z − z0 < R

−n

1 f ( z ) dz where an = , ∫ n +1 2πi c ( z − z0 ) 1 f ( z ) dz bn = , ∫ −n +1 2πi c ( z − z0 ) C is any positively oriented simple closed contour around z 0 and lying in the puctured disc 0 < z - z 0 < R.

∞

(ii ) ∑bn ( z − z0 )

−n

is called

n =1

principal part (PP) of the Laurent series, i.e. ∞

PP = ∑bn ( z − z0 )

−n

n =1

b1 b2 = + + ....... 2 z − z0 ( z − z0 )

If b k ≠ 0, for some k, say k = m, and b n = 0 ∀ n > m, then bm b1 b2 PP = + + ... + m z - z 0 ( z − z0 ) ( z − z0 )

Then the singularity z = z 0 of f(z) is called POLE OF ORDER m. If m = 1, then z 0 is a pole of order 1 and is called a SIMPLE POLE.

(iii) If an analytic function f(z) has a singularity other than a pole, then this singularity is known as ESSENTIAL SINGULARITY of f(z) , i.e.

if bn ≠ 0 for infinitely many n, then the singularity z 0 is called ESSENTIAL SINGULARITY of f(z).

(iv) If b n = 0

∀n,

then the singularity z 0 is called REMOVABLE SINGULARITY of f(z).

RESIDUE : The PP of the Laurent series is given by ∞

PP = ∑bn ( z − z0 )

−n

,

n =1

1

f ( z )dz bn = ∫ −n +1 2π i C ( z − z0 )

where

If n =1, then 1 b1 = 2πi

∫

f ( z ) dz

c

is called RESIDUE of f(z) at z = z 0 , and we write

b1 = Re s f ( z ) z = z0

1 = coeff of z − z0

Residue Theorem : Let C be a positively oriented simple closed contour. Suppose that f(z) is analytic within and on C except for a finite number of singular points z k (k = 1, 2,....n) inside C.

Then

∫

C

n

  f ( z )dz = 2π i ∑  Re s f ( z )  z = zk   k =1

How to find residue of a given fn f(z) : sin z Ex1 : Let f(z) = 4 , 0 < z < ∞. z 1 Now f(z) = 4 ( sin z ) z 3 5 7   1 z z z = 4  z − + − + .... (3)! (5)! (7)! z  

1 1 1 1 1 3 f ( z) = 3 − . + .z − z + .... (7)! z (3)! z (5)! 0< z <∞ 1 1 1 PP = − . + 3 (3)! z z Note that z = 0 is a pole of order ???

Hence 1 1 Res f ( z ) = b1 = coeff of = − z =0 z 6 sin z ∴ ∫ dz = 2 π i Res f ( z ) 4 z =0 z c: z =1

πi =− 3

Ex 2. Find the residue of f ( z ) = exp(1 / z ), and hence evaluate

∫ c

f(z)dz, C : z = 1.

Soln :  1 f ( z ) = exp    z 1 1 1 1 1 = 1+ + + + ..... 2 3 z 2! z 3! z

Note : z = 0 is an essential singularity of f(z). 1 ⇒ b1 = coeff of z = Re s f ( z ) z =0

=1

Hence

∫ c

f(z)dz = 2πi.

Ex 3. Find the residue of 2

f ( z ) = exp(1 / z ), and hence evaluate

∫ c

f(z)dz, C : z = 1.

Hints:

1. z = 0 is an essential singularity of f(z). 2. b1 = Re s f ( z ) = 0. z =0

3. I = 0.

How to find the residues ? We have ∞

∞

f(z) = ∑ an ( z − z0 ) + ∑ bn ( z − z0 ) n= 0

n

n =1

−n

Case IA : Let z = z 0 is a simple pole of f(z). Then ∞

f(z) = ∑ an ( z − z0 ) n =0

n

b1 + z − z0

⇒ ( z − z0 ) f ( z ) ∞

= b1 + ( z − z0 ) ∑ an ( z − z0 ) n= 0

⇒ lim ( z − z0 ) f ( z ) = b1 z → z0

= Re s f ( z ) z = z0

n

CaseIB : Let f(z) has a simple pole at z = z 0 and f(z) is of the form p( z ) f ( z) = , q( z ) where

(i) p(z) & q(z) are analytic at z = z 0 , (ii) p(z 0 ) ≠ 0, and (iii ) q( z ) has a simple zero at z = z 0 ,

Then p( z ) Re s f ( z ) = Re s z = z0 z = z0 q ( z ) p ( z0 ) = q′( z0 )

CaseII : Let z 0 be a pole of order m > 1 for the function f(z). ∞

Then f(z) = ∑ an ( z − z0 )

n

n =0

bm b1 b2 + + + .... + 2 m ( z − z0 ) ( z − z0 ) ( z − z0 )

⇒ ( z − z0 )

m

f ( z)

= ( z − z0 )

m

+ b1 ( z − z0 )

∞

∑ an ( z − z 0 )

n

n=0 m −1

+ b2 ( z − z0 )

+ ..... + bm−1 ( z − z0 ) + bm

m− 2

Let φ (z) = (z - z 0 )

m

f ( z)

then Res z =z 0

f ( z ) = b1 = coeff . of (z - z 0 )

m −1

in the

expansion of φ (z) =

φ

(m −1)

( z0 )

(m −1)!

by Taylor's Thm

Thus if z 0 is a pole of order m > 1 of f(z), then Res z = z0

φ

( m −1)

( z0 ) f ( z) = (m − 1)! 1 lim m −1 ( ) = φ z z → z0 (m − 1)!

[

]

Res z = z0

f ( z)

 1 lim  d m ( ) ( ) = z − z f z  z → z0  m − 1 0 (m − 1)! dz   m− 1

Ex1. Find the residue of f(z) at z = 0 and z = − 1, where 1 f(z) = . 2 z+z

Soln : Note that z = 0 and z = − 1 are simple poles of f(z).

∴

Res z= 0

f

lim ( z ) = z→ 0

lim = z→ 0

( z − 0) f ( z )  1    =1  1+ z 

∴

Res f z = −1

lim ( z ) = z→ 0

lim = z → −1

( z + 1) f ( z )  1   = − 1.  z

−z

e Q.2 (a) Evaluate I = ∫ dz . 2 z c: z =3 Soln : Clearly, z = 0 is a pole of order 2 of

−z

e f ( z) = 2 . z

Now I =

∫

f ( z )dz

c: z =3

= 2πi ∑ Re s f ( z ), z = zk

−z

e f ( z) = 2 z

∴

Re s z= 0

(

)

1 lim  d 2  f ( z) = . z→ 0  z f ( z)  (2 − 1)!  dz  =

lim  z→ 0 

d −z e   dz 

⇒

Re s z=0

f ( z) =

lim z→ 0

= −1 ∴ I = − 2π i

(− e ) −z

Q.2 (b) Evaluate −z

e I= ∫ dz . 2 z c: z −3 =1 Ans :

I = 0 (WHY ???)

Ex2(c). Evaluate I=

−z

e dz

∫ ( z −1) c: z =3

. 2

So ln : z = 1 is pole of order 2 of −z

e f ( z) = . 2 ( z −1)

Res ∴ z =1

( )

d −z f ( z) = e dz = −e

2πi ∴I = − e

−z

z =1

z =1

1 =− e

(c ) I =

∫

1 2 z z .e

dz

z =3

Let f(z)

1 2 z =z e

⇒ z = 0 is an essential singularity of f(z)

2

1 1 1 1 1 1 1  f ( z ) = z  1 + + . 2 + . 3 + 4 + ...  z 2! z 3! z 4! z  1 1 1 1 1 2 = z + z + + . + . 2 + .... 2! 3! z 4! z

Re s 1 1 ∴ f ( z ) = coeff . of = z=0 z 6 1 πi ∴ I = 2π i × = 6 3

(d ) z +1 I= ∫ 2 dz z − 2 z z =3 z +1 z +1 Let f(z) = 2 = z − 2 z z ( z − 2) ⇒ z = 0 & z = 2 are simple poles

Re s f ( z ) = lim z f ( z ) z →0 z =0 z +1 = lim z →0 z − 2 1 =− 2

lim Re s f ( z ) = ( z − 2) f ( z ) z =2 z →2 3 = 2 ∴ I = 2π i ∑ Re s f ( z )  1 3 = 2πi − +  = 2πi.  2 2

Q.3, p.233 Let f ( z ) be analytic at z 0 , and consider f ( z) g ( z) = . z − z0 Then Show that

(a ) If f(z 0 ) ≠ 0, then z0 is a simple pole of g(z) and Re s g(z) = f(z 0 )

z =z0

(b) If f(z 0 ) = 0, then z0 is a removable singularity of g(z) and Re s g(z) = 0. z = z0

Sol :

f(z) is analytic at z 0

⇒ f(z) has Taylor's series expansion about z 0 , &

f ( z ) = f(z 0 ) + ( z − z0 ) f ′( z0 ) ′ ′ f ( z0 ) 2 + ( z − z0 ) 2! 3 f ′′′( z0 ) + ( z − z0 ) + ... 3!

f ( z) ⇒g ( z ) = z − z0 f ( z0 ) = + f ′( z0 ) z − z0 f ′′( z0 ) + ( z − z0 ) 2! 2 f ′′′( z 0 ) + ( z − z0 ) +... 3!

(a) Clearly if f ( z0 ) ≠ 0, Then principal part (P.P) of g(z) is f ( z0 ) = z − z0

∴ z0 is a simple pole of g(z) and 1 Res g ( z ) = b1 = coeff of z = z0 z − z0 = f ( z0 )

(b) If f ( z0 ) = 0, then p.p.of g(z) is 0 ⇒ bn = 0 ∀ n ⇒ z = z0 is a removable singularity of g(z), and Res z = z0

g ( z) = 0

3

3z + 2 Q.4 (a) I = ∫ , c : z − 2 = 2 2 ( ) z − 1 z + 9 c

(

3

)

3z + 2 Let f ( z ) = 2 ( z − 1) z + 9

(

)

Then 1, 3i, − 3i are simple poles of f(z)

2i

2 -2i

4

Note : z = 1 is only inside C  3z + 2  Res ∴ f ( z) =  2  z =1  z +9  3

5 1 = = 10 2 Re s ∴ I = 2πi ×z =1 f ( z ) = πi

z =1

(b) c : z = 4 Then 1,3i ,−3i are all inside C 1 Re s ∴ z =1 f ( z ) = 2 4i

3i

1

-4 3i

4

Re s z =3i

3

3z + 2 f ( z) = ( z − 1)( z + 3i ) − 81i + 2 = ( 3i − 1)( 6i ) 2 − 81i = − 18 − 6i

z =3i

3

3z + 2 Re s f ( z ) = z =−3i ( z − 1)( z − 3i ) + 81i + 2 = ( − 3i − 1)( − 6i ) 2 + 81i = − 18 + 6i

z =−3i

∴ ∑ Res f ( z )

1 2 + 81i 2 − 81i = + − 2 6i − 18 6i + 18 =3

∴ I = 2π i ∑ Res f ( z ) = 6π i

dz Q.5 (b) I = ∫ 3 ,c : z + 2 = 3 ( ) z z + 4 c Let

1 f ( z) = 3 z ( z + 4)

-5

-4

-3

-2

-1

1

⇒ z = 0 is a pole of order 3 and z = −4 is a simple pole & both lie inside C.

2

Re s

1 d  1  ∴ f ( z) = . 2   z=0 2 dz  z + 4  1 = 3 4

z= 0

Re s

1 f (z ) = 3 z = −4 z

z = −4

1 =− 3 4

 1 1 ∴I = 2πi 3 − 3  = 0 4 4 