COMPLEX INTEGRATION
PART-B
1. State Cauchy’s integral formula.
Solution:
If f(z) is analytic inside and on a simple closed curve C and ‘a’ be any
point inside C then  f (z) dz  2i.f (a) where the integration being taken in
c
za
the positive direction around C.
2
2. Evaluate  z  1 dz where c is
(z  2)(z  3)
z 1
C
Solution:
Cauchy’s integral formula is  f (z) dz  2i.f (a) [ a lies inside C]
za
c
a=2 and 3 lies outside the circle
By Cauchy’s integral formula
z 1.
z2  1
 (z  2)(z  3)dz =0
C
2
3. Evaluate  3z  7z  1dz where c is
(z  3)
z 2
C
Solution:
The pole is at z=3 lies outside the circle
By Cauchy’s integral formula

C
z 2.
3z 2  7z  1
dz =0
(z  3)
2
4. Evaluate 
dz where c is
(z  1)(z  3)
z 1  2
C
Solution:
z  1  2 is a circle whose centre is 1 and radius 2.
z =-3 lies outside the circle z  1  2
z=1 lies inside the circle z  1  2
2

dz 
(z  1)(z  3)
C
 2 
 z  3
1


 (z  1) dz  2if (1)  2i  2   i
C
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COMPLEX INTEGRATION
5. Evaluate  (z 2  2z)dz where c is
z 1
C
Solution: Let
z  e i
dz  ieid
2
 (z  2z)dz 
2
 [(e
c
i 2
)  2ei ]ieid
0
2
i
 e

3i
 2ei2 d
0
2
 e 3i 2ei2 

 i

2i 
 3i
0
 e 3i


 ei2 
 3

2
0
2
 cos 3  i sin 3 
2

   cos 2  i sin 2 0
3

0
1
 (1  0)  (1  0)  [(1  0)  (1  0)]
3
0

z 2  2z
[or]
 (z
2
is analytic inside and on C.
 2z)dz
=0 by Cauchy’s integral formula.
C
6. Evaluate  (z  3)4dz where c the circle is
z3 4
C
Solution: Let
z  3  4ei
z  4ei  3
dz  4ieid
  (z  3) dz 
4
C
2


 4
5
  4e
4
i
 4ieid
0
  4
0
5
2
5 i5
e
id
cos5  i sin 502
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COMPLEX INTEGRATION
7. Evaluate  dz where c is the circle
z4
z  2.
C
Let f  z   1 and
z  4
lies outside of the circle
By cauchy’s integral formula, we have 

C
z
8. Evaluate  e dz where c is the circle
z1
z
1
 1.
2
z
1
 1.
2
C
Let f  z   ez and
z  1
lies outside of
z 2
dz
0.
z4
z
By cauchy’s integral formula, we have  e dz  0 .
z1
C
9. Evaluate 
C
dz
z3
Let f  z   1 and

where c is
z  3
z  1.
lies outside of the circle
z  1.
By cauchy’s integral formula, we have  dz  0
z3
C
10. Evaluate 
dz
C  z  3

f  z
C  z  
2
2

C  z  3
2
z  1.
 2if '  a 
a=3 lies outside
dz
where c is the circle
z  1.
0
11. Evaluate  cos z dz if c is
z 1
z 2
C
z 1
lies inside the circle
z 2
w.k.t cauchy’s integral formula is 
f  z
C  z  
= 2i  1
 2if  a 
a  1, f  z   cos z 


f  a   cos a 



f  1  cos   1

= 2i
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COMPLEX INTEGRATION
12. Using cauchy’s integral formula evaluate 
cos z 2
C  z  1 z  2 
dz
where c is
z 
3
2
z  1 lies inside the circle
z  2 lies inside the circle

the given integral can be re-written as 
cos z 2
z  2 dz
z 1
cos z 2
z2
cos  1
f  1 

1
1
1
f z 
By Cauchy’s integral formula 
cos z 2
z  2 dz  2i  1  2i
z 1
13. Expand log(1+z) Taylor’s series about z=0.
Solution:
f (z)  log(1  z)  f (0)  log1  0
1
f (z) 
 f (0)  1
1 z
1
f (z)  
 f (0)  1
(1  z)2
2(1  z)
2
f (z) 

 f (0)  2
4
(1  z)
(1  z)3
The Taylor series of f(z) about the point z=0 is given by
f (z)  f (0) 
z
z2
z3
f (0) 
f (0) 
f (0)  .....
1!
2!
3!
 0
z z2
z3

2
 ...
1! 2!
3!
z
z2 z3

 ...
2
3
14. Expand
1
z2
at z=1 in a Taylor’s series.
Solution:
1
 f (1)  1
z2
1
f (z)  
 f (1)  1
 z  22
f (z) 
f (z) 
2(z  2)

2
 f (1)  2
 z  2  z  23
The Taylor series of f(z) about the point z=0 is given by
4
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COMPLEX INTEGRATION
 z  a  f (a)  .....,here(a  1)
za
f (a) 
1!
2!
2
f (z)  f (a) 
 z  1 f (1)  .....
z 1
f (1) 
1!
2!
2
 f (1) 
 z  1 2 ( 2)  ...
 1  (z  1)( 1) 
2
2
 1  (z  1)  (z  1)  ......
15. Expand
z 1
z1
in Taylor’s series about z=1.
Solution:
f (z) 
f (z) 
z 1
 f (1)  0
z1
 z  1   z  1
f (z) 
 z  1
2
2  2  z  1 
 z  1
4


2
 z  1
4
 z  1
So the Taylor’s series for
 f (1) 
2
3
2 1

4 2
 f (1)  
z 1
z1
4
1

8
2
about z=1 is
 z  a  
za
f (a) 
f (a)  .....,here(a  1)
1!
2!
2
f (z)  f (a) 
 z  1 
z 1
f (1) 
f (1)  .....
1!
2!
2
 f (1) 
1
 1   z  1
 0    (z  1)   
 ...
2
2
 
 2
2
1
16. Find Laurent’s series of f (z)  z 2e z about z=0.
Solution:
Clearly f(z) is not analytic at z=0.
1
f (z)  z 2e z
2


1 1


z z
 z 2 1        .... 
1!
2!




z 1
 z 2    ....
1! 2!
17. Obtain the Laurent expansion of the function
ez
in the neighbourhood of
 z  1 2
its singular point. Hence find the residue at the point.
Solution:
Here z=1 is a singular point.
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COMPLEX INTEGRATION
f (z) 
ez
(z  1)2

e z  1 1
(z  1)2
ez 1 .e1

(z  1)2
 (z  1) (z  1)2

1 

 .....
2
1!
2!

(z  1) 
 1

1
1
 e1 

  .....
2
(z  1) 2
 (z  1)

e1

Residue at the point z=1 is the coefficient of
 Resf (z),1  e
1
z 1
18. State Cauchy’s Residue theorem.
If f(z) be analytic at all points inside and on a simple closed curve C,except
for a finite number of isolated singularities z1 ,z 2 ,.....,zn inside C,then
 f (z)dz  2i[sum of the residues of f (z) at z1 ,z 2 ,....,zn ]
c
n
 2i  R i where R i is the residue of f (z) at z  z i
i 1
19. Calculate the residue of f z  
f z  
e 2z
z  12
at the pole
e 2z
z  1
Given
Z=-1 is a pole of order 2
2
f z z 1 
Res
d 
e 2z 
z  12

z1 dz 
z  12 
lim
 
d 2z
e
z1 dz
 lim
 lim 2e 2z
z 1
2
 2e
z2
20. Find the residue of f z  
z  12 z  2
at z=-2.
Z=1 is a pole of order 2
Z=-2 is a pole of order 1
f z z  2 
Res

lim z  2
z 2
 lim


z  12 z  2
z2
z2
z  2 z  12

 22
 2  12

4 2

9 3
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COMPLEX INTEGRATION
f z  
z 1
z z  2
21. Find the residue of
Z=0 is a pole of order 1
Z=2 is a pole of order 1
z 1
z0 zz  2
Re sf z z 0  lim z
Re sf z z  2  lim z  2
z 2
z 1
3

zz  2 2
Z=2 is a simple pole
Re sf z z  2  lim z  2
z 2
4
z
3
z  2

1
2
f z  
22. Obtain the residues of the function
Z=-1 is a pole of order 1
Z=-2 is a pole of order 1
z3
z  1z  2 at its poles.
z  3  4
z  1z  2
z1
z  3  5
Re sf z z  2  lim z  2
z  1z  2
z2
Re sf z z  1  lim z  1
23. Consider the function f z  
f z  
sin z
z4


sin z
z4
.Find the pole and its order.

1 
z3 z5
z 

 .......... 
3
5
z 4 


1  z2 z4
1 

 .......... .
3
5

z 3 
Z=0 is a pole of order 3
24. Define pole and simple poles.
A point z=a is said to be a pole f(z) of order n if we can find a positive
lim z  an f z   0
za
integer n such that
A pole of order one is called a simple pole.
f z  
1
z  42 z  33 z  1
Example :
Hence z=1 is a simple pole of order 1
Z=4 is a simple pole of order 2
Z=3 is a simple pole of order 3.
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COMPLEX INTEGRATION
25. Find the regularities of the function f z  
f z  
Given
cot z
z  a3

cot z
z  a 3
cos z
sin z z  a 3
Singular points are poles and are given by Dr=0.
sin z  0,
z  a 3  0  z  a is a sin gular pole of order 3
sin z  sin n sin n
wheren  0,1,2,........
z  n  n  
z  n  0,1,2,........ are simple poles .
26. Find the principal part and residue at the pole of
f z  
2z  3
z  22
 ( 2z  3)( z  2)  2
[since principal part is negative powers]
( z  2) 2  z  2 is a sin gular pole of order 2
( 2 z  3) 
d 
( z  2) 2 .
2
z 2 dz 
( z  2) 2 
dz
where C is the square with vertices (0,0),(1,0),(1,1) and (0,1).
z2
Re sf ( z )z  2  Lim
27. Evaluate

c
Given
 z2
dz
c
Heref ( z )  1
a  2 lies outside
 ByCauchy ' sin tegralform ula
 z  2  0.
dz
c
*************************
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COMPLEX INTEGRATION
28. Find the residue of tan z at z 
f ( z )  tan z 

2
sin z P( z )

cos z Q( z )

P (a ) 
 Re sf ( z )z  a 


Q ' (a ) 

a
2
P( z )  sin z
Q( z )  cos z


P   sin  1 Q ' ( z )   sin z
2
 2


Q '     sin   1
2
 
 2
Re sf ( z )z   
2
P (a )
'
Q (a )

1
 1
1
29. Find a contour C for which
ez
 (z  1) 2 (z  1)dz
C
The poles of f(z) are given by z  1, z  1
z  1 is a simple pole
z  1is a pole of order 2

 e1 e
ez

Re sf ( z )z 1  Lim ( z  1) .

4
z 1 
( z  2) 2 ( z  1)  4
Re sf ( z )z  1  Lim
d
z  1 dz


ez
( z  1) 2 .


( z  1) 2 ( z  1) 
d  e z ( z  1)  e z

z  1 dz  ( z  1) 2
 Lim
 e 1 [2  1]  3


4
4e


 f ( z )dz  2i(sum of the residues )
C
 4e 2  12 
e 3 

 2i     2i 
 16e 
 4 4e 
30. Evaluate  sec zdz, where C is the unit circle
z  1.
C
Solution:
1
 sec zdz   cos z dz
C
C
Singular points of the function are got by equating the denominator to zero.
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COMPLEX INTEGRATION
cos z  0

z  (2n  1) ,n  0, 1, 2,...
2
 3
  ,  ,....
2
2
points lie outside z  1.
All
By Cauchy’s integral theorem
 f (z)dz  0
C
 sec zdz  0
C
31. Find the residue of
f (z) 
50z
(z  4)(z  1)2
at z=1.
Solution:
Z=1 is apole of order 2.
Resf (z)z 1 
 lim
(z  4)50  50z
z 1

d
50z
(z  1)2 .
z 1 dz
(z  4)(z  1)2
lim
(z  4)2
5 * 50  50
52

250  50 200

8
25
25
32. Give the formula to obtain the residue of f(z) that has a pole of order m at
z=a.
Solution:
If z  z 0 is a pole f order m,a simple formula to determine the residues is
given by
33. Find the
dm1 
(z  z0 )m f (z)
2 z z
m1 

(m  1)
0 dz
singular points of f (z)  1 1 state
sin
za
b1 
1
lim
their nature.
Solution:
f(z) has an finite number of poles which are given by
1
 n ,n  1, 2,....
za
1
i.e z  a 
n
1
z a
n
But z=a also is a singular point.
It is an essential singularity
It is a limit point of the poles
So it is a non isolated singularity.
SNSCT-DEPT OF MATHEMATICS
Page 10
COMPLEX INTEGRATION
SNSCT-DEPT OF MATHEMATICS
Page 11