proof of Cauchy residue theorem
Being f holomorphic by Cauchy-Riemann equations the differential form
f(z)dz is closed. So by the lemma about closed differential forms on a simple connected domain we know that the integral ∫Cf(z)𝑑z is equal to ∫C′f(z)𝑑z if C′ is any curve which is homotopic to C.
In particular we can consider a curve C′ which turns around the points aj along small circles and join these small circles with segments. Since the curve C′ follows each segment two times with opposite orientation it is enough to sum
the integrals of f around the small circles.
So letting z=aj+ρeiθ be a parameterization of the curve around the point aj, we have dz=ρieiθdθ and hence
∫Cf(z)𝑑z=∫C′f(z)𝑑z=∑jη(C,aj)∫∂Bρ(aj)f(z)𝑑z |
=∑jη(C,aj)∫2π0f(aj+ρeiθ)ρieiθ𝑑θ |
where ρ>0 is chosen so small that the balls Bρ(aj) are all disjoint and all contained in the domain U. So by linearity, it is enough to prove that for all j
i∫2π0f(aj+eiθ)ρeiθ𝑑θ=2πiRes(f,aj). |
Let now j be fixed and
consider now the Laurent series for f in aj:
f(z)=∑k∈ℤck(z-aj)k |
so that Res(f,aj)=c-1. We have
∫2π0f(aj+eiθ)ρeiθ𝑑θ=∑k∫2π0ck(ρeiθ)kρeiθ𝑑θ=ρk+1∑kck∫2π0ei(k+1)θ𝑑θ. |
Notice now that if k=-1 we have
ρk+1ck∫2π0ei(k+1)θ𝑑θ=c-1∫2π0𝑑θ=2πc-1=2πRes(f,aj) |
while for k≠-1 we have
∫2π0ei(k+1)θ𝑑θ=[ei(k+1)θi(k+1)]2π0=0. |
Hence the result follows.
Title | proof of Cauchy residue theorem |
---|---|
Canonical name | ProofOfCauchyResidueTheorem |
Date of creation | 2013-03-22 13:42:04 |
Last modified on | 2013-03-22 13:42:04 |
Owner | paolini (1187) |
Last modified by | paolini (1187) |
Numerical id | 7 |
Author | paolini (1187) |
Entry type | Proof |
Classification | msc 30E20 |