residue.htm Residue Theorem and Contour Integrals

Residues at Isolated Singularities
If z0 is a complex number we shall use the notation Cr(z0) for the closed circular path
Cr(z0)= {z | |z−z0| = r}
traversed in the counterclockwise direction.



Residue Theorem - One Singularity Version. Let f(z) be analytic in the region {z | 0 < |z − z0| < R }. Then for z inside Cr(z0), z ≠ z0,
f(z)
= ∞
∑
n = −∞ 
 an (z − z0)n,
an
= 1
2πi
 ⌠
(⎜)
⌡


Cr(z0) 
 f(ζ) (ζ−z0)−n − 1  dζ.
Here r is any number such that 0 < r < R.
In particular
a−1 = 1
2πi
 ⌠
(⎜)
⌡


Cr(z0) 
 f(ζ) dζ,
is the residue of f(z) at z=z0. so that
⌠
(⎜)
⌡


Cr(z0) 
 f(ζ) dζ = 2πi Resf(z)|z = z0.
 The Residue Theorem
Let C be a simple closed path. Suppose that f(z) is analytic on and inside C, except for a finite number of isolated singularities, z1, z2, …, zK inside C.
Then, by using cuts from C to small circles of small radius r around each zk,

respic1.gif


⌠
(⎜)
⌡


C 
 f(ζ)  dζ
= K
∑
k=1 
 ⌠
(⎜)
⌡


Cr(zk) 
 f(ζ) dζ
= K
∑
k=1 
 2πi Resf(z)|z = zk
= 2πi K
∑
k=1 
  Resf(z)|z = zk.


The Residue Theorem. Let C be a simple closed path.Suppose that f(z) is analytic on and inside C, except for a finite number of isolated singularities, z1, z2, …,zK inside C. Then
⌠
(⎜)
⌡


C 
 f(ζ) dζ = 2πi K
∑
k=1 
  Resf(z)|z = zk.


The Residue Theorem reduces the problem of evaluating a contour integral - an integral on a simple closed path - to the algebraic problem of determining the poles and residues1 of a function.


Exercises
Note the following special cases:

1.
Let C be a simple closed path. Suppose that f(z) is analytic on and inside C. Use the Residue Theorem to show that
⌠
(⎜)
⌡


C 
 f(ζ)  dζ = 0.
We knew this result already!

2.
Let C be a simple closed path. Let z be a point inside C. Find
⌠
(⎜)
⌡


C 
 1
ζ− z
  dζ.

3.
Let C be a simple closed path. Let z be a point outside C. Find
⌠
(⎜)
⌡


C 
 1
ζ− z
  dζ.

4.
Let C be a simple closed path. Let a and b be points inside C, a ≠ b. Find
⌠
(⎜)
⌡


C 
 1
(ζ− a)(ζ− b)
  dζ.


Finding the Residue

If f(z) has a pole of order M at z = z0, then
f(z)
= a−M
(z−z0)M
 + …+ a−1
z−z0
 + …,
(z−z0)M f(z)
= a−M + a−M + 1(z−z0)+ …+ a−1(z−z0)M−1+ ….
Since (z−z0)M f(z) is analytic at z = z0,
a−1 = 1
(M−1)!
 dM−1
dzM−1
 (z−z0)M f(z)|z = z0.

Partial Fractions


To find the partial fraction expansion of
f(z)
= qM + N − 1(z)
(z−z0)M pN(z)
 ,
degree qM + N − 1
= M+N−1,
degree pN
= N,
qM + N − 1(z0)
≠ 0,
pN(z0)
≠ 0,
let
h(z) = qM + N − 1(z)
pN(z)
 .
Near z=z0,
f(z)
= h(z0)
0!
 1
(z−z0)M
 + h′(z0)
1!
 1
(z−z0)M−1
        + …+ h(M−1)(z0)
(M−1)!
 1
z − z0
 + analytic.
Similar expansions may be found near the roots of pN(z) = 0.

Footnotes:

1Since functions behave so badly at an essential singularity, there is little hope of finding the residue at an essential singularity.

File translated from TEX by TTH, version 4.03.
On 22 Feb 2012, 13:38.