Section 8
SECTION 8
Residue Theory
(1) The Residue
(2) Evaluating Integrals using the Residue
(3) Formula for the Residue
(4) The Residue Theorem
1
What is a Residue?
Section 8
The residue of a function is the coefficient of the term
1
z  z0
in the Laurent series expansion (the coefficient b1).
f ( z )  a0  a1 ( z  z0 )  a2 ( z  z0 ) 2  a3 ( z  z0 )3  
b3
b1
b2
b4





2
3
4
z  z0 ( z  z0 ) ( z  z0 ) ( z  z0 )
Examples:
 2z  3
1 1 1 z z2
  2     
2
z  3z  2
z
z 2 4 8
1
2
 2  4z  12   8z  12   
1 z
b1  1
b1  0
2
What is a Residue?
Section 8
The residue of a function is the coefficient of the term
1
z  z0
in the Laurent series expansion (the coefficient b1).
f ( z )  a0  a1 ( z  z0 )  a2 ( z  z0 ) 2  a3 ( z  z0 )3  
b3
b1
b2
b4





2
3
4
z  z0 ( z  z0 ) ( z  z0 ) ( z  z0 )
Examples:
 2z  3
1 1 1 z z2
  2     
2
z  3z  2
z
z 2 4 8
1
2
 2  4z  12   8z  12   
1 z
b1  1
b1  0
3
What’s so great about the Residue?
Section 8
The formula for the coefficients of the Laurent series says
that (for f (z) analytic inside the annulus)
f ( z )  a0  a1 ( z  z0 )  a2 ( z  z0 ) 2  a3 ( z  z0 )3  
b3
b1
b2
b4





2
3
4
z  z0 ( z  z0 ) ( z  z0 ) ( z  z0 )
1
f ( z)
an 
dz,
n 1

2j C ( z  z0 )
So
1
n 1
bn 
f
(
z
)(
z

z
)
dz
0

2j C
 f ( z )dz  2jb
1
z0
C
C
We can use it to evaluate integrals
4
What’s so great about the Residue?
Section 8
The formula for the coefficients of the Laurent series says
that (for f (z) analytic inside the annulus)
f ( z )  a0  a1 ( z  z0 )  a2 ( z  z0 ) 2  a3 ( z  z0 )3  
b3
b1
b2
b4





2
3
4
z  z0 ( z  z0 ) ( z  z0 ) ( z  z0 )
1
f ( z)
an 
dz,
n 1

2j C ( z  z0 )
So
1
n 1
bn 
f
(
z
)(
z

z
)
dz
0

2j C
 f ( z )dz  2jb
1
z0
C
C
We can use it to evaluate integrals
5
Example (1)
Section 8
1
Integrate the function
counterclockwise about z 2
1 z
1  z  z 2  z 3   z  1
1

 1 1 1
1  z   2  3   1  z
z
 z z
1
C 1  z dz  2jb1  2j
z 2
singular
centre point
By Cauchy’s Integral Formula:
f ( z)
1
C z  z0 dz  2j f ( z0 )   C z 1 dz  2j f (1)  2j
6
Section 8
1  z  z 2  z 3   z  1
1

 1 1 1
1  z   2  3   1  z
z
 z z
z 2
singular
centre point
7
Example (1)
Section 8
1
Integrate the function
counterclockwise about z 2
1 z
1  z  z 2  z 3   z  1
1

 1 1 1
1  z   2  3   1  z
z
 z z
1
C 1  z dz  2jb1  2j
z 2
singular
centre point
By Cauchy’s Integral Formula:
f ( z)
1
C z  z0 dz  2j f ( z0 )   C z 1 dz  2j f (1)  2j
8
Example (1) cont.
Section 8
We could just as well let the centre be at z1
1
1
f ( z) 

,
1 z
z 1
0  z 1  
z 2
- a one-term Laurent series
1
C 1  z dz  2jb1  2j
centre /
singular
point
- as before
9
Example (2)
Section 8
 2z  3
Integrate the function z 2  3z  2 counterclockwise about z 3
3 5
9 2


z

z 
z 1

2 4
8

 2z  3
1 1 1 z z2

   2       1  z  2
2
z  3z  2 
z
z 2 4 8
  2  3  5  9 
2 z 
2
3
4

z z
z
z
 2z  3
C z 2  3z  2 dz  2jb1  2j
By Cauchy’s Integral Formula:
0
z  3/ 2
f ( z)
1
1
C z  z0 dz  2j f ( z0 )  C z  2 dz  C z 1 dz  2j f (1)  210j
Example (2)
Section 8
 2z  3
Integrate the function z 2  3z  2 counterclockwise about z 3
3 5
9 2


z

z 
z 1

2 4
8

 2z  3
1 1 1 z z2

   2       1  z  2
2
z  3z  2 
z
z 2 4 8
  2  3  5  9 
2 z 
2
3
4

z z
z
z
 2z  3
C z 2  3z  2 dz  2jb1  2j
By Cauchy’s Integral Formula:
0
z  3/ 2
f ( z)
1
1
C z  z0 dz  2j f ( z0 )  C z  2 dz  C z 1 dz  2j f (1)  211j
Section 8
So the Residue allows us to evaluate integrals of analytic
functions f (z) over closed curves C when f (z) has one singular
point inside C.
 f ( z )dz  2jb
1
z0
C
C
b1 is the residue of f (z) at z0
12
Section 8
That’s great - but every time we want to evaluate an integral
do we have to work out the whole series ?
No - in the case of poles - there’s a quick and easy way
to find the residue
We’ll do 3 things:
e.g. 4 sin z
1. Formula for finding the residue for a simple pole
z 1
3z  3
e.g. ( z  1) 2
2. Formula for finding the residue for a pole of order 2
3. Formula for finding the residue for a pole of
2e z
any order
e.g. (3z  j ) 7
13
Formula for finding the residue for a simple pole
Section 8
If f (z) has a simple pole at z0, then the Laurent series is
f ( z )  a0  a1 ( z  z0 )   
b1
z  z0
0  z  z
0
 R
 ( z  z0 ) f ( z )  a0 ( z  z0 )  a1 ( z  z0 ) 2    b1
 lim ( z  z0 ) f ( z)  b1
z  z0
we’re putting the centre at
the singular point here
Res f ( z )  lim ( z  z0 ) f ( z )
z  z0
z  z0
14
Formula for finding the residue for a simple pole
Section 8
If f (z) has a simple pole at z0, then the Laurent series is
f ( z )  a0  a1 ( z  z0 )   
b1
z  z0
0  z  z
0
 R
 ( z  z0 ) f ( z )  a0 ( z  z0 )  a1 ( z  z0 ) 2    b1
 lim ( z  z0 ) f ( z)  b1
z  z0
we’re putting the centre at
the singular point here
Res f ( z )  lim ( z  z0 ) f ( z )
z  z0
z  z0
15
Formula for finding the residue for a simple pole
Section 8
If f (z) has a simple pole at z0, then the Laurent series is
f ( z )  a0  a1 ( z  z0 )   
b1
z  z0
0  z  z
0
 R
 ( z  z0 ) f ( z )  a0 ( z  z0 )  a1 ( z  z0 ) 2    b1
 lim ( z  z0 ) f ( z)  b1
z  z0
we’re putting the centre at
the singular point here
Res f ( z )  lim ( z  z0 ) f ( z )
z  z0
z  z0
16
Example (1)
Section 8
2z  j
Find the residue of f ( z )  ( z  j )( z 2  1) at zj
Res f ( z )  lim ( z  z0 ) f ( z )
z  z0
z  z0
( z  j )( 2 z  j )
(2 z  j )
j
 lim
 lim

z i ( z  j )( z 2  1)
z i ( z  j ) 2
4
Check: the Laurent series is
f ( z) 

2z  j
2( z  j )  j  1  2( z  j )  j 
1




2
2
( z  j )( z 2  1)
z j
(
z

j
)
z

j


2
j

(
z

j
)




 2( z  j )  j 

2( z  j )  j 
1
( z  j)
( z  j)2
( z  j )3

3
4
 


1  2
( z  j )( 2 j ) 2  1  ( z  j ) /( 2 j ) 2  ( z  j )( 2 j ) 2 
2j
(2 j ) 2
(2 j )3

i 1
1 5
1

  ( z  j)  ( z  j)2 
0 z j 2
4 z  j 4 16
2


17
Example (2)
Find the residue at the poles of f ( z ) 
Section 8
z 1
z 2  2z
z 1
z 1
1
Res f ( z )  lim z
 lim

z 0
z 0
z ( z  2) z 0 z  2
2
z 1
z 1 3
Res f ( z )  lim ( z  2)
 lim

z 2
z

2
z 2
z ( z  2)
z
2
Check: the Laurent series are

z 1
z 1  1 
z 1  z z 2
1 3 3z 3z 2
f ( z) 




1   2       
z ( z  2)
2 z 1  z / 2 
2z  2 2
2 z 4 8 16

 ( z  2)  3 

z 1
( z  2)  3 
1
1
f ( z) 





z ( z  2)
( z  2)  2  ( z  2)  2( z  2) 1   ( z  2) / 2
0  z  2
0  z  2  2

( z  2)  3  z  2 ( z  2) 2
3
1 ( z  2) ( z  2) 2


  
 


1 
2
2( z  2) 
2
2
4
8
 2( z  2) 2
18
Example (2)
Find the residue at the poles of f ( z ) 
Section 8
z 1
z 2  2z
z 1
z 1
1
Res f ( z )  lim z
 lim

z 0
z 0
z ( z  2) z 0 z  2
2
z 1
z 1 3
Res f ( z )  lim ( z  2)
 lim

z 2
z

2
z 2
z ( z  2)
z
2
Check: the Laurent series are

z 1
z 1  1 
z 1  z z 2
1 3 3z 3z 2
f ( z) 




1   2       
z ( z  2)
2 z 1  z / 2 
2z  2 2
2 z 4 8 16

 ( z  2)  3 

z 1
( z  2)  3 
1
1
f ( z) 





z ( z  2)
( z  2)  2  ( z  2)  2( z  2) 1   ( z  2) / 2
0  z  2
0  z  2  2

( z  2)  3  z  2 ( z  2) 2
3
1 ( z  2) ( z  2) 2


  
 


1 
2
2( z  2) 
2
2
4
8
 2( z  2) 2
19
Example (2)
Find the residue at the poles of f ( z ) 
Section 8
z 1
z 2  2z
z 1
z 1
1
Res f ( z )  lim z
 lim

z 0
z 0
z ( z  2) z 0 z  2
2
z 1
z 1 3
Res f ( z )  lim ( z  2)
 lim

z 2
z

2
z 2
z ( z  2)
z
2
Check: the Laurent series are

z 1
z 1  1 
z 1  z z 2
1 3 3z 3z 2
f ( z) 




1   2       
z ( z  2)
2 z 1  z / 2 
2z  2 2
2 z 4 8 16

 ( z  2)  3 

z 1
( z  2)  3 
1
1
f ( z) 





z ( z  2)
( z  2)  2  ( z  2)  2( z  2) 1   ( z  2) / 2
0  z  2
0  z  2  2

( z  2)  3  z  2 ( z  2) 2
3
1 ( z  2) ( z  2) 2


  
 


1 
2
2( z  2) 
2
2
4
8
 2( z  2) 2
20
Example (2)
Find the residue at the poles of f ( z ) 
Section 8
z 1
z 2  2z
z 1
z 1
1
Res f ( z )  lim z
 lim

z 0
z 0
z ( z  2) z 0 z  2
2
z 1
z 1 3
Res f ( z )  lim ( z  2)
 lim

z 2
z

2
z 2
z ( z  2)
z
2
Check: the Laurent series are

z 1
z 1  1 
z 1  z z 2
1 3 3z 3z 2
f ( z) 




1   2       
z ( z  2)
2 z 1  z / 2 
2z  2 2
2 z 4 8 16

 ( z  2)  3 

z 1
( z  2)  3 
1
1
f ( z) 





z ( z  2)
( z  2)  2  ( z  2)  2( z  2) 1   ( z  2) / 2
0  z  2
0  z  2  2

( z  2)  3  z  2 ( z  2) 2
3
1 ( z  2) ( z  2) 2


  
 


1 
2
2( z  2) 
2
2
4
8
 2( z  2) 2
21
Example (2)
Find the residue at the poles of f ( z ) 
Section 8
z 1
z 2  2z
z 1
z 1
1
Res f ( z )  lim z
 lim

z 0
z 0
z ( z  2) z 0 z  2
2
z 1
z 1 3
Res f ( z )  lim ( z  2)
 lim

z 2
z

2
z 2
z ( z  2)
z
2
Check: the Laurent series are

z 1
z 1  1 
z 1  z z 2
1 3 3z 3z 2
f ( z) 




1   2       
z ( z  2)
2 z 1  z / 2 
2z  2 2
2 z 4 8 16

 ( z  2)  3 

z 1
( z  2)  3 
1
1
f ( z) 





z ( z  2)
( z  2)  2  ( z  2)  2( z  2) 1   ( z  2) / 2
0  z  2
0  z  2  2

( z  2)  3  z  2 ( z  2) 2
3
1 ( z  2) ( z  2) 2


  
 


1 
2
2( z  2) 
2
2
4
8
 2( z  2) 2
22
Question:
Section 8
z2  3
Find the residue at the pole z01 of f ( z ) 
z ( z  1)
23
Formula for finding the residue for a pole of order 2
Section 8
If f (z) has a pole of order 2 at z0, then the Laurent series is
f ( z )  a0  a1 ( z  z0 )   
b1
b2

z  z0 ( z  z0 ) 2
 ( z  z0 ) 2 f ( z )  a0 ( z  z0 ) 2  a1 ( z  z0 )3    b1 ( z  z0 )  b2
now differentiate:
d
 ( z  z0 ) 2 f ( z )  2a0 ( z  z0 )  3a1 ( z  z0 ) 2    b1
dz
d

 lim
( z  z0 ) 2 f ( z )  b1
z  z dz
0

d
2
Res f ( z )  lim
( z  z0 ) f ( z )
z  z0 dz
z  z0

24
Example
Section 8
z
Find the residue of f ( z )  ( z  2)( z  1) 2

at z1

d
Res f ( z )  lim
( z  z0 ) 2 f ( z )
z  z 0 dz
z  z0
 2  2
d  z 
 lim
 lim 


2


z 1 dz z  2

 z 1  ( z  2)  9
Check: the Laurent series is
f ( z) 
 ( z  1)  1 

z
( z  1)  1 
1
1






( z  2)( z  1) 2
( z  1) 2  3  ( z  1)  3( z  1) 2 1  (( z  1) / 3) 
 ( z  1)  1  1
( z  1)  1  z  1 ( z  1) 2
1
1 ( z  1) 

1







 3 


2 
2
2
2
3( z  1) 
3
3
3
3
 ( z  1) 3( z  1) 3



1
2
2 2( z  1)




2
3( z  1) 9( z  1) 27
81
0  z 1  3
25
Formula for finding the residue for a pole of any order
Section 8
If f (z) has a pole of order m at z0, then the Laurent series is
bm
b1
b2
f ( z )  a0  a1 ( z  z0 )   


2
z  z0 ( z  z0 )
( z  z0 ) m
 ( z  z0 ) m f ( z )  a0 ( z  z0 ) m  a1 ( z  z0 ) m1    b1 ( z  z0 ) m1
 b2 ( z  z0 ) m2    bm
now differentiate m1 times and let zz0 to get:


d ( m1)
 lim ( m1) ( z  z0 ) m f ( z )  (m  1)!b1
z  z0 dz

1
d ( m1)
Res f ( z ) 
lim ( m1) ( z  z0 ) m f ( z )
z  z0
(m  1)! z  z0 dz

26
The Residue Theorem
Section 8
We saw that the integral of an analytic function f (z) over a
closed curve C when f (z) has one singular point inside C is
 f ( z )dz  2jb
1
z0
C
C
b1 is the residue of f (z) at z0
Residue Theorem: Let f (z) be an analytic
function inside and on a closed path C
except for at k singular points inside C.
Then
k
C
 f ( z)dz  2j  Res f ( z)
C
i 1
z  zi
27
Section 8
Example
2 z
Integrate the function 2
around z  2
z z
2 z
2 z
2 z 

 Res 2

2
C z 2  z dz  2j Res
z 0 z  z
z 1 z  z

C
2 z
2 z
Res 2
 lim
 2
z 0 z  1
z 0 z  z
2 z
2 z
Res 2
 lim
3
z 1
z 1 z  z
z
2 z
  2 dz  2j
z z
C
28
Section 8
Example
2 z
Integrate the function 2
around z  2
z z
2 z
2 z
2 z 

 Res 2

2
C z 2  z dz  2j Res
z 0 z  z
z 1 z  z

C
2 z
2 z
Res 2
 lim
 2
z 0 z  1
z 0 z  z
2 z
2 z
Res 2
 lim
3
z 1
z 1 z  z
z
2 z
  2 dz  2j
z z
C
29
Section 8
Example
2 z
Integrate the function 2
around z  2
z z
2 z
2 z
2 z 

 Res 2

2
C z 2  z dz  2j Res
z 0 z  z
z 1 z  z

C
2 z
2 z
Res 2
 lim
 2
z 0 z  1
z 0 z  z
2 z
2 z
Res 2
 lim
3
z 1
z 1 z  z
z
2 z
  2 dz  2j
z z
C
30
Section 8
Example
2 z
Integrate the function 2
around z  2
z z
2 z
2 z
2 z 

 Res 2

2
C z 2  z dz  2j Res
z 0 z  z
z 1 z  z

C
2 z
2 z
Res 2
 lim
 2
z 0 z  1
z 0 z  z
2 z
2 z
Res 2
 lim
3
z 1
z 1 z  z
z
2 z
  2 dz  2j
z z
C
31
Section 8
Example
2 z
Integrate the function 2
around z  2
z z
2 z
2 z
2 z 

 Res 2

2
C z 2  z dz  2j Res
z 0 z  z
z 1 z  z

C
2 z
2 z
Res 2
 lim
 2
z 0 z  1
z 0 z  z
2 z
2 z
Res 2
 lim
3
z 1
z 1 z  z
z
2 z
  2 dz  2j
z z
C
32
Proof of Residue Theorem
Section 8
Enclose all the singular points
with little circles C1, C1,  Ck.
C
f (z) is analytic in here
By Cauchy’s Integral Theorm for multiply connected regions:
 f ( z)dz   f ( z )dz   f ( z)dz     f ( z)dz
C
C1
C2
Ck
But the integrals around each of the small circles is just the
residue at each singular point inside that circle, and so
k
 f ( z)dz  2j  Res f ( z)
C
i 1
z  zi
33
Section 8
Topics not Covered
(1) Another formula for the residue at a simple pole (when f (z)
is a rational function p(z)q(z),
Res f ( z ) 
z  z0
p ( z0 )
q( z0 )
(2) Evaluation of real integrals using the Residue theorem
2
e.g.

0
d
2  sin 
using
z  e j
(3) Evaluation of improper integrals using the Residue theorem

e.g.
x2 1
 x 4  5 x 2  4 dx
34