proof of Cauchy integral formula
Let $$ be a disk in the complex plane, $S\subset D$ a finite subset, and $U\subset \u2102$ an open domain that contains the closed disk $\overline{D}$. Suppose that

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$f:U\backslash S\to \u2102$ is holomorphic, and that

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$f(z)$ is bounded^{} near all $z\in D\backslash S$.
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
$$g(\zeta )=\frac{f(\zeta )f(z)}{\zeta z},\zeta \in D\backslash {S}^{\prime},$$ 
where ${S}^{\prime}=S\cup \{z\}$. Note that $g(\zeta )$ is holomorphic and bounded on $D\backslash {S}^{\prime}$. The second assertion is true, because
$$g(\zeta )\to {f}^{\prime}(z),\text{as}\zeta \to z.$$ 
Therefore, by the Cauchy integral theorem
$${\oint}_{C}g(\zeta )\mathit{d}\zeta =0,$$ 
where $C$ is the counterclockwise circular contour parameterized by
$$\zeta ={z}_{0}+R{e}^{it},\mathrm{\hspace{0.33em}0}\le t\le 2\pi .$$ 
Hence,
$${\oint}_{C}\frac{f(\zeta )}{\zeta z}\mathit{d}\zeta ={\oint}_{C}\frac{f(z)}{\zeta z}\mathit{d}\zeta .$$  (1) 
$\mathrm{\mathbf{L}\mathbf{e}\mathbf{m}\mathbf{m}\mathbf{a}}$ If $z\in \u2102$ is such that $\parallel z\parallel \ne 1$, then
$$ 
The proof is a fun exercise in elementary integral calculus, an application of the halfangle trigonometric substitutions.
Thanks to the Lemma, the right hand side of (1) evaluates to $2\pi if(z).$ Dividing through by $2\pi i$, we obtain
$$f(z)=\frac{1}{2\pi i}{\oint}_{C}\frac{f(\zeta )}{\zeta z}\mathit{d}\zeta ,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}},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}^{\prime}(z)=\frac{1}{2\pi i}\frac{d}{dz}{\oint}_{C}\frac{f(\zeta )}{\zeta z}\mathit{d}\zeta =\frac{1}{2\pi i}{\oint}_{C}\frac{f(\zeta )}{{(\zeta z)}^{2}}\mathit{d}\zeta ,z\in D.$$ 
Title  proof of Cauchy integral formula 

Canonical name  ProofOfCauchyIntegralFormula 
Date of creation  20130322 12:47:23 
Last modified on  20130322 12:47:23 
Owner  rmilson (146) 
Last modified by  rmilson (146) 
Numerical id  15 
Author  rmilson (146) 
Entry type  Proof 
Classification  msc 30E20 