ProofOfCauchyResidueTheorem 2003-06-18 12:23:18 2006-07-25 01:50:09 Proof Cauchy residue theorem 0 % 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} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 chosen 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.