proof of Cauchy integral formula

Let be a disk in thecomplex plane, $S\subset D$ a finite subset, and $U\subset ℂ$ anopen domain that contains the closed disk $\overline{D}$. Suppose that

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

• $f(z)$ is bounded near all $z\in D\backslash \backslash S$.

Hence, by a straightforward compactness argument we also have that$f(z)$ is bounded on $\overline{D}\backslash \backslash S$, and hence bounded on$D\backslash \backslash S$.

Let $z\in D\backslash \backslash S$ be given, and set

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

where $S^{\prime }=S\cup \{z\}$. Note that $g(\zeta )$ is holomorphic andbounded on $D\backslash \backslash S^{\prime }$.The second assertion is true, because

$$g(\zeta )\to f^{\prime }(z),\text{as}\zeta \to z.$$

Therefore, by the Cauchy integral theorem

$${\oint }_{C}g(\zeta )𝑑\zeta =0,$$

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,

$${\oint }_C\frac{f(\zeta )}{\zeta -z}𝑑\zeta ={\oint }_C\frac{f(z)}{\zeta -z}𝑑\zeta .$$(1)

$\mathrm{𝐋𝐞𝐦𝐦𝐚}$If $z\in ℂ$ is such that $\parallel z\parallel \ne 1$, then

The proof is a fun exercise in elementary integral calculus, anapplication 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

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

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 uniformconvergence, the order of the derivative and the integral operationscan be interchanged. In this way we obtain the second formula:

$$f^{\prime }(z)=\frac{1}{2\pi i}\frac{d}{dz}{\oint }_C\frac{f(\zeta )}{\zeta -z}𝑑\zeta =\frac{1}{2\pi i}{\oint }_C\frac{f(\zeta )}{{(\zeta -z)}^2}𝑑\zeta ,z\in D.$$