4The calculus of residues

IB Complex Methods

4.1 The residue theorem
We are now going to massively generalize the last result we had in the previous
chapter. We are going consider a function
f
with many singularities, and obtain
an analogous formula.
Theorem
(Residue theorem)
.
Suppose
f
is analytic in a simply-connected
region except at a finite number of isolated singularities
z
1
, ··· , z
n
, and that a
simple closed contour γ encircles the singularities anticlockwise. Then
I
γ
f(z) dz = 2πi
n
X
k=1
res
z=z
k
f(z).
z
1
z
2
z
3
γ
Note that we have already proved the case
n
= 1 in the previous section. To
prove this, we just need a difficult drawing.
Proof.
Consider the following curve
ˆγ
, consisting of small clockwise circles
γ
1
, ··· , γ
n
around each singularity; cross cuts, which cancel in the limit as they
approach each other, in pairs; and the large outer curve (which is the same as
γ
in the limit).
ˆγ
z
1
z
2
z
3
Note that
ˆγ
encircles no singularities. So
H
ˆγ
f
(
z
) d
z
= 0 by Cauchy’s theorem.
So in the limit when the cross cuts cancel, we have
I
γ
f(z) dz +
n
X
k=1
I
γ
k
f(z) dz =
I
ˆγ
f(z) dz = 0.
But from what we did in the previous section, we know
I
γ
k
f(z) dz = 2πi res
z=z
k
f(z),
since
γ
k
encircles only one singularity, and we get a negative sign since
γ
k
is a
clockwise contour. Then the result follows.
This is the key practical result of this course. We are going to be using this
very extensively to compute integrals.