16.7 The Residue Theorem
The following is the residue theorem.
Theorem 16.7.1 Let Ω be an open set and let γk :
,m, be a parametrizaton for a
simple closed curve, and be continuous and of bounded variation. Suppose also that
for all z
. Then if f
: Ω → X has every singularity removable or a pole and such that no γk∗ contains
any poles of f,
where here P 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 singularities α which are not in the unbounded
component of ℂ ∖∪k=1mγk∗. Thus there exists a finite set,
such that these are the
only possibilities for which ∑
might not equal zero. Therefore,
and it is this last equation which is established. For z near αj,
where gj is analytic at and near αj. Now define
It follows that G
has a removable singularity at each
. Therefore, by Corollary 16.4.3
The above is certainly a grand and glorious result but we typically have in mind something much more
ordinary. As a review of the main ideas consider this.
You have a simple closed curve oriented such that the winding number is 1. Say γ is a parametrization
for this curve. Then inside there are finitely many singularities
. Enclose each with a circle
oriented in the clockwise direction, parameterized by
and connect them with straight lines as
shown. Then you have two simple closed curves which intersect in these finitely many straight
line segments. Orient them oppositely so that line integrals over the straight line segments
cancel and each of the two simple closed curves is oriented positively. Then if f
on the inside of γ
except for the ak
and is continuous on γ∗,
the Cauchy integral theorem
Letting γk ≡−
Now on the inside of γk,
because all the other terms have primitives. Indeed, if n≠−
as a primitive. However,
In words, the contour integral is 2πi times the sum of the residues. This formulation is in a sense more
general because it is only required that the function be continuous on γ∗ and analytic on the inside of γ∗
except for the exceptional points indicated.
So is there a way to find the residues? The answer is yes.
Procedure 16.7.2 Say you want to find res
This is the case where you have a pole of order M at a. You would multiply by
M. This would
Then you would take M − 1 derivatives and then take the limit as z → a. This would give
You can see from the formula that this will work and so there is no question that the limit exists.
Because of this, you could use L’Hospitals rule to formally find this limit. This rule pertains only to real
functions of a real variable. However, since you know the limit exists in this case, you can pick a one
dimensional direction and apply L’Hospital to the real and imaginary parts to identify the limit which is
typically what needs to be done.