ProofOfCauchyIntegralFormula 2002-06-13 14:01:41 2006-09-11 10:26:05 Proof Cauchy integral formula 0 \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} \newcommand{\nl}[1]{{\PMlinkescapetext{{#1}}}} \newcommand{\pln}[2]{{\PMlinkname{{#1}}{#2}}} Let $D=\{z\in\cnums:\Vert z-z_0\Vert< R\}$ be a disk in the complex plane, $S\subset D$ a finite subset, and $U\subset\cnums$ an open domain that contains the closed disk $\overline{D}$. Suppose that \begin{itemize} \item $f:U\backslash S\rightarrow \cnums$ is holomorphic, and that \item $f(z)$ is bounded near all $z\in D\backslash S$. \end{itemize} Hence, by a straightforward compactness argument we also have that $f(z)$ is bounded on $\overline{D}\backslash S$, and hence bounded on $D\backslash S$. Let $z\in D\backslash S$ be given, and set $$g(\zeta) = \frac{f(\zeta)-f(z)}{\zeta-z},\quad \zeta\in D\backslash S',$$ where $S'=S\cup \{ z\}$. Note that $g(\zeta)$ is holomorphic and bounded on $ D\backslash S'$. The second assertion is true, because $$g(\zeta)\rightarrow f'(z),\;\mbox{as}\; \zeta\rightarrow z.$$ Therefore, by the Cauchy integral theorem $$ \oint_C g(\zeta)\, d\zeta=0, $$ where $C$ is the counterclockwise circular contour parameterized by $$\zeta = z_0 + R e^{it},\; 0\leq t\leq 2\pi.$$ Hence, \begin{equation} \label{eq:1} \oint_C \frac{f(\zeta)}{\zeta-z}\, d\zeta = \oint_C \frac{f(z)}{\zeta-z}\, d\zeta. \end{equation} \noindent $\mathbf{Lemma}$ If $z\in\cnums$ is such that $\Vert z\Vert\neq 1$, then $$ \oint_{\Vert\zeta\Vert=1} \frac{d\zeta}{\zeta-z} = \begin{cases} 0 &\text{if } \Vert z\Vert>1\\ 2\pi i &\text{if } \Vert z\Vert<1\\ \end{cases} $$ The proof is a fun exercise in elementary integral calculus, an application of the half-angle trigonometric substitutions. Thanks to the Lemma, the right hand side of \eqref{eq:1} evaluates to $2\pi i f(z).$ Dividing through by $2\pi i$, we obtain $$ f(z) = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{\zeta-z}\, d\zeta, \quad z\in D, $$ as desired. Since a circle is a compact set, the defining limit for the derivative $$\frac{d}{dz} \frac{f(\zeta)}{\zeta-z}= \frac{f(\zeta)}{(\zeta-z)^2},\quad z\in D$$ converges uniformly for $\zeta\in \partial D$. Thanks to the uniform convergence, the order of the derivative and the integral operations can be interchanged. In this way we obtain the second formula: $$ f'(z) = \frac{1}{2\pi i} \frac{d}{dz} \oint_C \frac{f(\zeta)}{\zeta-z}\, d\zeta = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta,\quad z\in D.$$