Cauchy integral formula

Let \(\Omega\subseteq\CC\) be open, let \(f\) be holomorphic on \(\Omega\), and let \(z_0\in\Omega\) and \(R>0\) satisfy \(\overline{D_R(z_0)}\subseteq\Omega\). Let \(\gamma\coloneqq\partial D_R(z_0)\), oriented counterclockwise. Then for every \(n\geq0\), \[ f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\gamma\frac{f(\xi)}{(\xi-z_0)^{n+1}}\dxi, \] and in particular \(f(z_0)=\frac{1}{2\pi i}\int_\gamma\frac{f(\xi)}{\xi-z_0}\dxi\). Consequently, the coefficients of the Taylor expansion \(f(z)=\sum_{k\geq0}c_k(z-z_0)^k\) at \(z_0\) are \[ c_k=\frac{f^{(k)}(z_0)}{k!}=\frac{1}{2\pi i}\int_\gamma\frac{f(\xi)}{(\xi-z_0)^{k+1}}\dxi. \]