# proof of Cauchy integral formula

Let $D=\{z\in\mathbb{C}:\|z-z_{0}\| be a disk in the complex plane, $S\subset D$ a finite subset, and $U\subset\mathbb{C}$ an open domain that contains the closed disk $\overline{D}$. Suppose that

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

 $g(\zeta)=\frac{f(\zeta)-f(z)}{\zeta-z},\quad\zeta\in D\backslash S^{\prime},$

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

 $g(\zeta)\rightarrow f^{\prime}(z),\;\mbox{as}\;\zeta\rightarrow z.$

Therefore, by the Cauchy integral theorem

 $\oint_{C}g(\zeta)\,d\zeta=0,$

where $C$ is the counterclockwise circular contour parameterized by

 $\zeta=z_{0}+Re^{it},\;0\leq t\leq 2\pi.$

Hence,

 $\oint_{C}\frac{f(\zeta)}{\zeta-z}\,d\zeta=\oint_{C}\frac{f(z)}{\zeta-z}\,d\zeta.$ (1)

$\mathbf{Lemma}$ If $z\in\mathbb{C}$ is such that $\|z\|\neq 1$, then

 $\oint_{\|\zeta\|=1}\frac{d\zeta}{\zeta-z}=\begin{cases}0&\text{if }\|z\|>1\\ 2\pi i&\text{if }\|z\|<1\\ \end{cases}$

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 (1) evaluates to $2\pi if(z).$ Dividing through by $2\pi i$, we obtain

 $f(z)=\frac{1}{2\pi i}\oint_{C}\frac{f(\zeta)}{\zeta-z}\,d\zeta,\quad z\in D,$

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:

 $f^{\prime}(z)=\frac{1}{2\pi i}\frac{d}{dz}\oint_{C}\frac{f(\zeta)}{\zeta-z}\,d% \zeta=\frac{1}{2\pi i}\oint_{C}\frac{f(\zeta)}{(\zeta-z)^{2}}\,d\zeta,\quad z% \in D.$
Title proof of Cauchy integral formula ProofOfCauchyIntegralFormula 2013-03-22 12:47:23 2013-03-22 12:47:23 rmilson (146) rmilson (146) 15 rmilson (146) Proof msc 30E20