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.