COMPLEX ANALYSIS 199

 1   1 1 1 
(b)  z  3 sin     z  3   3
 5
 .........
 z 2  z  2 ( z  2) .3! ( z  2) .5! 
so, z = -2 is an isolated essential singular point.

5.6 Residue of a Complex Function

Residue at a pole:
Let, z = a be a pole of order ‘m’ of f (z) and C1 is a circle of radius ‘r’ with center at z = a which does not
contain singularities except z = a, then f(z) is analytic within the annular region r < | z – a | < R can be expanded
into Laurrent series within the annulur region as:

R
r
 
a
c1
f  z   an ( z  a )n   bn ( z  a ) n
n 0 n 1

1
2 i C1
Co-efficient b1 is known as residue of f (z) at z = a i.e. Res. f(z = a) = b1 = f ( z )dz

Methods of finding residues:

CASE 1: Residue at simple pole:

(z) (a)
(b) Method 2: If f(z) = where (a) = 0 but (a)  0, then Res. f(z = a) =
(z) (a)

CASE 2: Residue at a pole of order ‘n’:

1  d n1 n 
(a) Method 1: Res. f (z = a) =  n1 ( z  a) f ( z) 
(n  1)!  dz z  a
(b) Method 2: First put z - a = t and expand it into series, then Res. f (z = a) = co-efficient of 1/t.

CASE 3: Residue at z =  :

Res. f  z     Lt   zf ( z )
z 
Example 30: Find the singular points of the following function and the corresponding residues:

1  2z z2
(a) f ( z )  (b) f ( z )  2 (c) f ( z )  z 2 e1/ z
z  z  1 z  2  z  a2
 200 COMPLEX ANALYSIS

Soln: (a) f  z   1  2z
 Poles : z  0, z  1, z  2
z  z  1  z  2 
1  2z 1
Res. f  z  0   Lt  z  0  f  z   Lt 
z 0 z0  z  1 z  2  2
1  2z
Res. f  z  1  Lt  z  1 f  z   Lt 1
z 1 z 1 z  z  2 

1  2z 3
Res. f  z  2   Lt  z  2  f  z   Lt 
z 2 z 2 z  z  1 2

z2
(b) f  z    Poles : z  ia, z  ia
z 2  a2
 z2  1  z2  1
Res. f  z  ia      ia; Res. f  z  ia       ia
 2 z  z ia 2  2 z  z ia 2

2 1/ z 2  1 1 1 
(c) f  z   z e  z 1   2  3  .......  Poles : z  0
 z z .2! z .3! 
1 1 1
Res. f  z  0   Coefficient of  
z 3! 6

sin z
Example 31: The value of the residue of is
z6
1 1 2 i 2 i
(a)  (b) (c) (d) 
5! 5! 5! 5!

sin z 1  z3 z5 z 7 
Soln: f z  6
 6 
z     ........
z z  3! 5! 7! 
1 1 1 11
    ......
z 5 3! z 3 5! z
1 1
Residue of f  z  = coefficient of   
 z  5!
Correct option is (b)

5.7 Cauchy Residue Theorem:

If f (z) in single-valued and analytic inside a simple closed curve ‘C, except at a finite number of singular
points within C, then

 f (z)dz  2i  sum of the residues at poles within 'C '

4  3z 3
Example 32: Evaluate the integral:  z  z  1 z  3 dz where z  2
C

4  3z
Soln: f  z    Poles : z  0, z  1, z  3
z  z  1 z  3
But, the given contour is circle centered at the origin and radius 3/2 units.
Therefore, only z = 0 and z = 1 within the contour.
 4 1  5 i
I  2 i  Re s. f  z  0   Re s. f  z  1   2 i    
3 2 3

e2 z  z 2
Example 33: Evaluate the integral:   z  1
C
5
dz where z  2

2z 2
Soln: f  z   e  z5  Poles : z  1 (order 5)
 z  1
1 d 4 2z 2 2e 2 4 ie2
I  2 i  Re s. f  z  1  2 i   e  z   2 i  
4! dz 4   z 1 3 3

z3
Example 34: The value of the integral  dz , where C is closed contour defined by the equation
C z 2  5z  6
2 | z | 5  0 , traversed in the anti-clockwise direction, is
(a) 16 i (b) 16 i (c) 8 i (d) 2 i

z3
Soln: f  z   ; Condition of singularity: z 2  5z  6  0  z  3, 2 1st order pole 
z 2  5z  6
5
Given closed contour: c : 2 z  5  0  z 
2
Only z = 2 i.e. point (2, 0) is within ‘c’

z3
Res.f  z  z 2   z  2   8
 z  3 z  2 z2
Given integral: I  2i   sum of the residues   16i
Correct answer is (a)
Example 35: The value of the integral  z10 dz , C is the unit circle with the origin as the centre is
C

11
(a) 0 (b) z / 11 (c) 2 iz11 / 11 (d) 1/1
Soln: f  z   z10 is analytic in the entire complex argand plane.
So, according to Cauchy integral theorem,
10
z dz  0
C

Correct option is (a)