proof of Cauchy integral formula

Let $D=\{z\in\mathbb{C}:\|z-z_{0}\|<R\}$ 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

•

$f:U\backslash S\rightarrow\mathbb{C}$ is holomorphic, and that
•

$f(z)$ 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},\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
Canonical name	ProofOfCauchyIntegralFormula
Date of creation	2013-03-22 12:47:23
Last modified on	2013-03-22 12:47:23
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	15
Author	rmilson (146)
Entry type	Proof
Classification	msc 30E20