Chapter 53

ResiduesDefinition 53.0.1 The residue of f at an isolated singularity α which is a pole, writtenres( f ,α) is the coefficient of (z−α)−1 where

f (z) = g(z)+m

∑k=1

bk

(z−α)k .

Thus res( f ,α) = b1 in the above.

At this point, recall Corollary 51.7.20 which is stated here for convenience.

Corollary 53.0.2 Let Ω be an open set and let γk : [ak,bk]→ Ω, k = 1, · · · ,m, be closed,continuous and of bounded variation. Suppose also that

m

∑k=1

n(γk,z) = 0

for all z /∈Ω. Then if f : Ω→ C is analytic,

m

∑k=1

∫γk

f (w)dw = 0.

The following theorem is called the residue theorem. Note the resemblance to Corollary51.7.20.

Theorem 53.0.3 Let Ω be an open set and let γk : [ak,bk]→ Ω, k = 1, · · · ,m, be closed,continuous and of bounded variation. Suppose also that

m

∑k=1

n(γk,z) = 0

for all z /∈Ω. Then if f : Ω→ Ĉ is meromorphic such that no γ∗k contains any poles of f ,

12πi

m

∑k=1

∫γk

f (w)dw = ∑α∈A

res( f ,α)m

∑k=1

n(γk,α) (53.0.1)

where here A denotes the set of poles of f in Ω. The sum on the right is a finite sum.

Proof: First note that there are at most finitely many α which are not in the unboundedcomponent of C \∪m

k=1γk ([ak,bk]) . Thus there exists a finite set, {α1, · · · ,αN} ⊆ A suchthat these are the only possibilities for which ∑

nk=1 n(γk,α) might not equal zero. There-

fore, 53.0.1 reduces to

12πi

m

∑k=1

∫γk

f (w)dw =N

∑j=1

res( f ,α j)n

∑k=1

n(γk,α j)

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