181202(MA 2161) – MATHEMATICS - II UNIT 4 notes




Unit. 4 Complex Integration
1
UNIT IV COMPLEX INTEGRATION
Part-A
Problem 1 Evaluate
 3 C 1
z dz
z   where C is z  2 using Cauchy’s integral formula
Solution:
Given
 3 C 1
z dz
z  
Here f z  z , a 1lies inside z  2
 
  3
2 1
C 1 2!
zdz i f
z
   
 
 i0  f 1  0
 3 0
C 1
zdz
z
 
  .
Problem 2 State Cauchy’s Integral formula
Solution:
If f z is analytic inside and on a closed curve C that encloses a simply
connected region R and if 'a ' is any point in R , then   1  
2 C
f z
f a dz
 i z a

  .
Problem 3 Evaluate
1
z
C
e dz whereC is z  2 1.
Solution:
1
ez is analytic inside and onC .
Hence by Cauchy’s integral theorem
1
z 0
C
e dz 
Problem 4 Classify the singularities of  
 
1
2
ez f z
z a


.
Solution:
Poles of f z are obtained by equating the denominator to zero.
i.e.,  2 z  a  0 , z  a is a pole of order 2
Unit. 4 Complex Integration
2
The principal part of the Laurent’s expansion of e1/ z about z = 0 contains infinite number
terms. Therefore there is an essential singularity at z = 0.
Problem 5 Calculate the residue of  
2
3
f z 1 e z
z

 at the poles.
Solution:
Given  
2
3
f z 1 e z
z


Here z  0 is a pole of order 3
   
2 2
3
0 2 3
Re 1 0 1
0 2!
z
z
Lt d e s f z z
 z dz z
  
        
2
2
2
1 1
2! 0
z Lt d e
z dz
    
1 2 2
2! 0
z Lt d e
z dz
   
1 4 2
2! 0
z Lt
e
z
 

1  4 2
2
    .
Problem 6 Evaluate cos
1 C
z dz
z

  if C is z  2 .
Solution:
We know that, Cauchy Integral formula is
    2
C
f z
dz if a
z a
 
  if 'a ' lies inside C
cos ,
1 C
z dz
z

  Here f  z  cos z
z 1lies inside C
 f 1  cos 1  1.
cos 2  1 2
1 C
z dz i i
z

      
  .
Problem 7 Define Removable singularity
Solution:
A singular point 0 z  z is called a removable singularity of f z is  
0
Lt
f z
z  z
exists
finitely
Unit. 4 Complex Integration
3
Example: For f  z sin z
z
 , z = 0 is a removable singularity since   1
0
Lt
f z
z


Problem 8 Test for singularity of 2
1
z 1
and hence find corresponding residues.
Solution:
Let   2   
1 1
1
f z
z z i z i
 
  
Here z  i is a simple pole
z  i is a simple pole
      
Lt 1
Res z i z i
z i z i z i
  
  
 
1 1
2
Lt
z i z i i
 
 
      
Lt 1
Res z i z i
z i z i z i
   
   
1
2i


.
Problem 9 What is the value of z
C
e dz where C is z 1.
Solution:
Put z  ei
dz  iei d
 
2
0
........... 1 z ei i
C
e dz e ie d 

    
Put t  ei dt  ei d
When  0, t 1,   2 , ssss t 1
(1)
1 1
1 1
z t t 0
C
e dz  e dt  e  
Problem 10 Evaluate
3 2 7 1
1 C
z z dz
z
 
  , where 1
2
z  .
Solution:
Given
3 2 7 1
1 C
z z dz
z
 
 
Here f  z  3z2  7z 1
z  1 lies outside 1
2
z 
Unit. 4 Complex Integration
4
Here
3 2 72 1 0
1 C
z dz
z
 

  .(By Cauchy Theorem)
Problem 11 State Cauchy’s residue theorem
Solution:
If f  z be analytic at all points inside and on a simple closed curveC , except for
a finite number of isolated singularities 1 2 , ,..., n z z z inside C
then     1 2 2 , ,..., n
C
 f z dz   isum of the residue of f z at z z z  .
Problem 12 Calculate the residue of  
 
2
2 1
e z f z
z


at its pole.
Solution:
Given  
 
2
2 1
e z f z
z


Here z  1is a pole of order 2
   
 
2
2
1 2
1 1
11! 1
z
z
Lt d e Resf z z
 z dz z
      
2 2 2 2
1
z Lt
e e
z
  
 
.
Problem 13 Using Cauchy integral formula evaluate   
cos 2 ,
1 2 C
z dz
z z

   where
3
2
z 
Solution:
    
cos 2 cos 2 cos 2
1 2 1 2 C C C
z dz z dz z dz
z z z z
   
 
      
    
1 , 1 1
1 2 1 2
A B A B
z z z z
 
            
Here f z  cos z2
z 1 lies inside 3
2
z 
z  2 lies outside 3
2
z 
Hence by Cauchy integral formula
Unit. 4 Complex Integration
5
    
cos 2 2
1 2 C
z dz i f z
z z

  
  
 2 i 1
 2 i [ f  z  cos z, f 1  cos  1]
Problem 14 State Laurent’s series
Solution:
If 1 C and 2 C are two concentric circles with centres at z  a and radii 1 r and 2 r
  1 2 r  r and if f  z is analytic on 1 C and 2 C and throughout the annular region R
between them, then at each point z in R ,
   
0 1  
n n
n n
n n
f z a z a b
z a
 
 
  
   ,
where
 
  1
1 , 0,1,2,...
n 2 n
C
f z dz
a n
 i z a   
  ,
 
  1
1 , 1, 2,3,...
n 2 n
C
f z dz
b n
 i z a    
 
Problem 15 Find the zeros of
3
3
1
1
z
z


.
Solution:
The zeros of f z are given by f z  0,
3
3
1 0
1
z
z



i.e.,
1
z3 1  0, z  (1)3
z 1, w,w2 (Cubic roots of unity)
Part-B
Problem 1 Using Cauchy integral formula evaluate
 2   C 1 2
dz
z  z   where C the
circle 3
2
z  .
Solution:
Here z  1 is a pole lies inside the circle
z  2 is a pole lies out side the circle
 2    2
1
2
C 1 2 1
dz z dz
z z z
  
    
Here f z 1
z 2


Unit. 4 Complex Integration
6
  2
1
( 2)
f z
z
  

Hence by Cauchy integral formula
 
 
  1
2
!
n
n
C
f z dz i f a
z a n

 
 
 2   2
1
2
C 1 2 C [ ( 1)]
dz z dz
z z z
 
     
2 ( 1)
1!
i f    
 2  2
2 1 1
1 2 2
i f z
z

                       
2 1
9
 i
      
2
9
 i

 .
Problem 2 Evaluate 2
( 1) C
z dz
z z

  where C is the circle z  3 .
Solution:
W.K.T   1  
2 C
f z
f a dz
 i z a

 
Given 2
( 1) C
z dz
z z

  Here z  0 , z 1lies inside the circle
Also f z  z  2
Now 1
( 1) 1
A B
z z z z
 
 
Put z  0 A  1
z 1 B 1
1 1 1
z(z 1) z z 1
  
 
2 2 2
( 1) 1 C C C
z dz z dz z dz
z z z z
  
  
    
 2 i f 0  2 i f 1
 2 i f (1)  f (0)
 2 i1 (2)
 2 i2 1  2 i .
Unit. 4 Complex Integration
7
Problem 3 Find the Laurent’s Series expansion of the function   
1
2 3
z
z z

 
, valid
in the region 2  z  3.
Solution:
Let     
1
2 3
f z z
z z


 
  
1
2 3 2 3
z A B
z z z z

 
   
z 1  A(z  3)  B(z  2)
Put z  2
2 1  A(2  3)  0
A  3
Put z  3
31  A0  B3 2
4  B
B  4
  3 4
2 3
f z
z z

  
 
Given region is 2  z  3
2  z and z  3
2 1
z
 and 1
3
z 
  3 4
1 2 3 1
3
f z
z z
z

  
         
   
1 1 3 1 2 4 1
3 3
z
z z
              
   
2 2 3 1 2 2 .... 4 1 ...
3 3 3
z z
z z z
                                
Problem 4 Expand     
7 2
2 1
f z z
z z z


 
valid in 1 z 1  3
Solution:
Given     
7 2
2 1
f z z
z z z


 
    
7 2
2 1 2 1
f z z A B C
z z z z z z

   
   
Unit. 4 Complex Integration
8
7z  2  A z  2 z 1  Bz z 1 Cz z  2
Put z  2
B  2
Put z  0
A 1
Put z  1
C  3
  1 2 3
2 1
f z
z z z
   
 
Given region is 1 z 1  3
Let u  z 11 u  3
z  u 11 u & u  3
1 1 & 1
3
u
u
  
  1 2 3
1 3
f z
u u u
   
 
1 2 3
(1 1) 3(1 )
3
u u u
u
  
  
1 1 1 1 1 2 1 3
3 3
u
u u u
              
   
2 2 1 1 1 1 ... 2 1 ... 3
3 3 3
u u
u u u u
                                
2 2 1 1 1 1 ... 2 1 1 1 ... 3
1 1 1 3 3 3 1
z z
z z z z
                                            
 
 
 
2 0
2 1 2 1
1 1 3 3
n
n n
n n
z
f z
z z
 
 

    
    .
Problem 5 Expand     
2 1
2 3
f z z
z z


 
as a Taylor series valid in the
region z  2.
Solution:
Given     
2 1
2 3
f z z
z z


 
Now  z  2 z  3  z2  5z  6
     
2 1 5 7 1
2 3 2 3
z z
z z z z
  
  
   
Unit. 4 Complex Integration
9
Now   
5 7
2 3 2 3
z A B
z z z z
 
 
   
5z  7  Az  3  Bz  2
Put z  2
A  3
Put z  3
B  8
  1 3 8
2 3
f z
z z
   
 
Given z  2
  1 3 8
2 1 3 1
2 3
f z
z z
  
         
   
1 1 1 3 1 8 1
2 2 3 3
z z               
   
2 2 1 3 1 ... 8 1 ...
2 2 2 3 3 3
    z   z       z   z                   
   
0 0
1 3 1 8 1
2 2 3 3
n n
n n
n n
 z  z
 
            
     
    1 1
0
1 1 3 8
2 3
n n
n n
n
f z z

 

         .
Problem 6 Using Cauchy Integral formula Evaluate
6
3
sin
6
C
z dz
z     
 
 where C is
circle z 1.
Solution:
Here f z  sin6 z
f z  6sin5 cos z
f  z  6 sin6 z  cos2 z.5sin4 z
Here
6
a 
 , clearly
6
a 
 lies inside the circle z 1
By Cauchy integral formula
 
 
  3
2
2! C
f z i f a
z a
  
 
Unit. 4 Complex Integration
10
6
3
sin 2
2! 6
6
C
z i f
z
 

     
      
 

6 sin6 5cos2 sin4
6 6 6
i   

                   
6 1 5 3
64 16 4
  i      
6 1 15
64 64
  i     
6 15 1
64
 i
       
21
16
 i

Problem 7 Expand f  z  sin z into a Taylor’s series about
4
z 
 .
Solution:
Given f  z  sin z
f z  cos z
f  z  sin z
f z  cos z
Here
4
a 

sin 1
4 4 2
f           
   
cos 1
4 4 2
f          
   
sin 1
4 4 2
f            
   
cos 1
4 4 2
f            
   
W.K.T Taylor’s series of f  z at z  a is
         
2
...
1! 2!
z a z a f z f a f a f a
       
 
2
4 4 ...
4 1! 4 2! 4
z z
f z f f f
 
  
                         
     
Unit. 4 Complex Integration
11
2
1 1 4 1 ...
2 4 2 2 2
z
z


                         
 
Problem 8 Evaluate  2 
sec
C 1
z z dz
 z  where C is the ellipse 4x2  9y2  9, using
Cauchy’s residue theorem.
Solution:
Equation of ellipse is
4x2  9y2  9
2 2
1
9 / 4 1
x  y 
i.e.,  
2 2
2 1
3 1
2
x  y 
Major axis is 3
2
, Minor axis is 1.
The ellipse meets the x axis at 3
2
 and the y axis at 1
Given   2
sec
1
f z z z
z


1 1 cos
z
z z z

 
The poles are the solutions of 1 z1 zcos z  0
i.e., z  1, z 1are simple poles and 2 1
2
z n 
 
Out of these poles z 1 lies inside the ellipse
, 3
4 4
z  
   lies outside the ellipse
  1      Re 1
z 1 1 1 cos
Lt z s f z z
 z z z z
      
11 cos
Lt z
z z z


 
1
2cos1


  1      Re 1
z 1 1 1 cos
Lt z s f z z
 z z z z
       
11 cos
Lt z
z z z

  
Unit. 4 Complex Integration
12
1 1
2cos1 2cos1
 
 
  2
sec 2 sum of the residues
1 C
z z dz i
z
  
 
2 1 1
2cos1 2cos1
 i
        
 2 isec1.
Problem 9 Using Cauchy integral formula evaluate (i) 2
4
2 5 C
z dz
z z

   , where C is
the circle z 1 i  2 (ii) 4 3
( 1)( 2) C
z dz
z z z

   , C is the circle 3
2
z  .
Solution:
(i) Given z 1 i  2
z  1 i  2 is a circle whose centre is 1 i and radius 2.
i.e., centre 1,1 and radius 2
z2  2z  5  z 1 2iz  (1 2i
1 2i i.e., 1,2lies inside the C
1 2i i.e., 1,2lies out side the C
2 2 5 0 2 4 20 , 1 2
2
z z z z i
  
           
 
   
4
1 2 1 2 C
z dz
z i z i


         

 
 
4
1 2
1 2 C
z
z i
dz
z i

     
   
   
Hence f z 4
1 2
z
z i


    
Here by Cauchy integral formula
    2
C
f z
dz i f a
z a
 
 
  2
4 2 1 2
2 5 C
z i f i
z z


  
  
   
2 1 2 4
1 2 1 2
i i
i i

    
         
Unit. 4 Complex Integration
13
2 3 2 3 2 .
4 2
i i i
i


         
(ii) 4 3
( 1)( 2) C
z dz
z z z

  
z  0, z 1 lie inside the circle 3
2
z 
z  2 lies outside the circle
4 3
( 1)( 2) 1 2
z A B C
z z z z z z

   
   
4 3z  A(z 1)(z  2)  B(z)(z  2) C(z)(z 1)
Put z  0
4  4A
A 1
Put z 1
B  1
Put z  2
C  1
4 3 2 1 1
( 1)( 2) 1 2
z
z z z z z z

   
   
4 3 2 1 1
( 1)( 2) 1 2 C C C C
z dz dz dz dz
z z z z z z

  
       
 
 
  1
2 0
!
n
n
C
f z i f
z a n

 
 
 2 [2 i f 0] 2 i f 1  0
 4 i f (0)  2 i f 1
 4 i (1)  2 i 1
 2 i  f 0 1 f 1 1
Problem 10 Using Cauchy’s integral formula evaluate  2 4 2 C
dz
z   where C is circle
z i  2
Solution:
 2 2  2  2
1 1
z 4 z 2i z 2i

  
Given z i  2 , centre 0,1 , radius 2
z  2i lies outside the circle
z  2i lies inside the circle
Unit. 4 Complex Integration
14
 
 
 
2
2 2 2
1
2
C 4 c 2
dz z i dz
z z i

 
   
Here  
 2
1
2
f z
z i


 
 3
2
2
f z
z i
  

 
 3  3
2 2 2
2 2 4
f i
i i i
    

2
64 32
  i   i
Hence by Cauchy Integral Formula
 
 
  1
2
!
n
n
C
f z i dz f a
z a n

 
 
 
 
  2 2
2 2
C 4 1! 16
f z i f i
z
    
  .
Problem 11 Find the Laurent’s series which represents the function
 1 2
z
z  z 
in (i) z  2 (ii) z 1 1
Solution:
(i). Let      1 2
f z z
z z

 
Now  1 2 1 2
z A B
z z z z
 
   
z  Az  2  Bz 1
Put z  1
A  1
Put z  2
B 1
  1 2
1 2
f z
z z

  
 
Given z  2, 2  z i.e., 2 1 1 1
z z
  
  1 2
1 2
f z
z z

  
 
Unit. 4 Complex Integration
15
1 2
z 1 1 z 1 2
z z

 
         
   
1 1 1 1 1 2 1 2
z z z z
              
   
(ii). z 1 1
Let u  z 1
i.e., u 1
  1 2
1 2
f z
z z

 
 
1 2
u 1 u

 

1 21 u 1
u
 
  
1 2(1 u u2 ...)
u

    
1 2 1 (1 ) (1 )2 ...
1
z z
z
            
Problem 12 Prove that 2 1 a 2
2
a 2a cos 1
2 d
0 


  
 

, given a2 < 1.
Solution: Let 

  


2
0 a 2a cos 1
I d 2
Put z = ei
Then
iz
d  dz and cos  = 



 
z
z 1
2
1
     

C
2 1 z
1 z
iz
dz
a a
I where C is |z| = 1.
   

C  
z2 1
dz
ai
1
z a
1 a
     

C
z2 1
dz
a
i
a z
a 1
 
C
f (z)dz where   z 1
1
a
f (z) i
z a
1 a 2   



 
   a
z a z 1
1
a
i
 



 
The singularities of f(z) are simple poles at a and
a
1 . a2 < 1 implies |a| < 1 and 1
a
1 
Unit. 4 Complex Integration
16
 The pole that lies inside C is z = a.
Res[f(z); a]    a
z a z a z 1
1
a
Lim(z a). i
   



  
  a
1 a
1
a
i




 
a 1
i
2 

Hence 2 1 a 2
2
a 1
I 2 i. i




 
Problem 13 Show that
5 4cos 6
2 cos 2 .d
0


 
  

Solution: Let 

 
 

2
0 5 4cos
I cos 2 .d
Put z = ei
Then
iz
d  dz and cos  = 



 
z
z 1
2
1


 


2 
0 5 4 cos
I Re al Part of e .d
i2
    

C
2
z 1
z
iz
. dz
5 2
z
Re al Part of where C is |z| = 1.
  

C
2
2
z z 1
z .dz
2i
Re al Part of 1
2 5
     

C
2
2 z 2
z 1
z .dz
2i
Re al Part of 1
 
C
Re al Part of f (z)dz where    2 z 2
z 1
z2 .
2i
f (z) 1
 

z = -½ and z = -2 are simple poles of f(z).
z = -½ lies inside C.
     2 z 2
z 1
2
2
1
2
- 1
2
1 . z
2i
Res f(z); - Lim (z ). 1
z  
 

12i
. 1
2i
1
2
3
4
1
 
 I = Real Part of 2i.
12i
1
Unit. 4 Complex Integration
17
O
ח
-R R X
Y
= Real Part of 6

= 6
 .
Problem 14 Prove that  2 2
0 1 4
dx
x
 

 
Solution:
Let  
 2 1 2 C C
z dz dz
z
 
  
Where  
 2 2
1
1
z
z
 

Here C is the semicircle  bounded by the diameter R, R
By Cauchy residue theorem,
      ......(1)
R
C R
 z dz  x dx  z dz
 
    
To evaluate of  
C
 z dz
The poles of  
 2 2
1
1
z
z
 

is the solution of  z2 1 2  0
i.e.,  2  2 z  i z i  0
i.e., the poles are z  i, z  i
z  i lies with inside the semi circle
z  i lies outside the semi circle
Now ( ) 1  2  
z i 1!
Lt d Res z z i z
z i dz
    

Unit. 4 Complex Integration
18
 
 
2
2 2
1 1
1! 1
Lt
z i
z i z
 
   
     
2   
2
1 1 1
1! ( )
Lt d z z i z i
z i dz z i
 
          

 3
Lt 2
z i z i


 
 3
2 2 1
i i 2i 4i
 
  

  2  at
C
 z dz   i Sum of residues of  z its poles which lies inC
2 1 ..........(2)
4 2
i
i

      
Let R, then z  so that  z  0
  0.........(3)
Lt
z dz
z


 
  
Sub (2) and (3) in (1)
   
C
 z dz  x dx


  
 2 1 2 2
dx
x
 

 
 
 2 2
0
2
1 2
dx
x
 
 
 
 2 2
0 1 4
dx
x
 
 
  .
Problem 15 Evaluate 2 2
0
x sin x dx
x a

 
Solution:
2 2 2 2
0
2 x sin x dx x sin x dx
x a x a
 


   
2 2 2 2
0
sin 1 sin
2
x x dx x x dx
x a x a
 


   
2 2
1 sin
2
z z dz
z a



 
Unit. 4 Complex Integration
19
1 ........(1)
2
 I
Now z sin z is the imaginary part of zeiz
2 2
z sin z dz
z a


 
 
2 2 . .
zeiz dz
z a


  
 
Let   2 2   
z eiz zeiz z
z a z ia z ia
  
  
The poles are z  ia , z  ia
Now the poles z  ia lies in the upper half – plane
But z  ia lies in the lower half – plane.
Hence
      
iz
z ia
Lt ze Res z z ia
z ia z ia z ia


      
 
Lt zeiz
z ia z ia

 
2 2
iae a e a
ia
 
 
  2 2 2 Sum of the residues at each poles in the upper half plane
zeiz dz i
z a



 
 
2
2
e a  i
  
  
 
 iea
I=I.P. of 2 2
zeiz dz
z a

  
= I.P. of  iea 
  ea ...........(2)
Sub (2) in (1)
2 2
0
sin 1 1
2 2
x x dx x e a
x a


  
 


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