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

• $f:U\backslash S\rightarrow \cnums$ is holomorphic, and that
• $f(z)$ is bounded near all $z\in D\backslash S$ .

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} $\mathbf{Lemma}$ If $z\in\cnums$ is such that $\Vert z\Vert\neq 1$ , then

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 () 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: $$ f'(z) = \frac{1}{2\pi i} \frac{d}{dz} \oint_C \frac{f(\zeta)}{\zeta-z}\, d\zeta = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta,\quad z\in D.$$