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'proof of Cauchy residue theorem'
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| Title of object: |
proof of Cauchy residue theorem |
| Canonical Name: |
ProofOfCauchyResidueTheorem |
| Type: |
Proof |
| Created on: |
2003-06-18 12:23:18 |
| Modified on: |
2003-06-18 12:27:01 |
| Classification: |
msc:30E20 |
Revision comment (for changes between this and next version):
| Changes for correction #8801 ('typo'). |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
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%\usepackage{xypic}
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% define commands here |
Content:
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 $\int_C f(z)\, dz$ is equal to $\int_{C'} f(z)\, dz$ if $C'$ is any curve which is homotopic to $C$.
In particular we can consider a curve $C'$ which turns around the points $a_j$ 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 $z=a_j+\rho e^{i\theta}$ be a parameterization of the curve around the point $a_j$, we have $dz=\rho i e^{i\theta}\, d \theta$ and hence
\[
\int_C f(z)\, dz = \int_{C'} f(z)\, dz = \sum_j \eta(C,a_j)\int_{\partial B_\rho(a_j)} f(z)\, dz
\]\[
= \sum_j \eta(C,a_j) \int_0^{2\pi} f(a_j+\rho e^{i\theta}) \rho i e^{i\theta}\, d\theta
\]
where $\rho>0$ is choosen so small that the balls $B_\rho(a_j)$ are all disjoint and all contained in the domain $U$. So by linearity, it is enough to prove that
for all $j$
\[
i\int_0^{2\pi} f(a_j+e^{i\theta})\rho e^{i\theta}\, d\theta = 2\pi i \mathrm{Res}(f,a_j).
\]
Let now $j$ be fixed and
consider now the Laurent series for $f$ in $a_j$:
\[
f(z)= \sum_{k\in \mathbb Z} c_k (z-a_j)^k
\]
so that $\mathrm{Res}(f,a_j)=c_{-1}$. We have
\[
\int_0^{2\pi} f(a_j+e^{i\theta})\rho e^{i\theta}\, d\theta =
\sum_k \int_0^{2\pi} c_k (\rho e^{i\theta})^k \rho e^{i\theta}\, d\theta
=\rho^{k+1} \sum_k c_k \int_0^{2\pi} e^{i(k+1)\theta}\, d\theta.
\]
Notice now that if $k=-1$ we have
\[
\rho^{k+1} c_k \int_0^{2\pi} e^{i(k+1)\theta}\, d\theta =
c_{-1}int_0^{2\pi} d\theta = 2\pi c_{-1} = 2\pi \,\mathrm{Res}(f,a_j)
\]
while for $k\neq -1$ we have
\[
\int_0^{2\pi} e^{i(k+1)\theta}\, d\theta = \left[\frac{e^{i(k+1)\theta}}{i(k+1)}\right]_0^{2\pi} = 0.
\]
Hence the result follows. |
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