Let \(\Omega\subseteq\CC\) be open, let \(f\colon\Omega\to\CC\), and let \(z_0\in\Omega\). The function \(f\) is complex differentiable, or holomorphic, at \(z_0\) if the limit \[ f'(z_0)\coloneqq\lim_{\substack{h\to0\\ h\in\CC\setminus\{0\}}}\frac{f(z_0+h)-f(z_0)}{h} \] exists. The function \(f\) is holomorphic on \(\Omega\) if it is holomorphic at every point of \(\Omega\), and a function \(f\colon\CC\to\CC\) that is holomorphic on \(\CC\) is entire.
Let \(\Omega\subseteq\CC\) be open, \(f\colon\Omega\to\CC\), and \(z_0\in\Omega\). Then \(f\) is holomorphic at \(z_0\) if and only if there exist \(\alpha\in\CC\) and a function \(\psi\), defined for \(h\in\CC\) with \(z_0+h\in\Omega\), such that \(\psi(h)\to0\) as \(h\to0\) and \[ f(z_0+h)-f(z_0)=\alpha h+h\,\psi(h). \] In this case \(\alpha=f'(z_0)\).
If \(f'(z_0)\) exists, put \(\alpha\coloneqq f'(z_0)\), \(\psi(0)\coloneqq0\), and \(\psi(h)\coloneqq\frac{f(z_0+h)-f(z_0)}{h}-\alpha\) for \(h\neq0\); then \(\psi(h)\to0\) and the identity holds. Conversely, dividing the identity by \(h\neq0\) gives \(\frac{f(z_0+h)-f(z_0)}{h}=\alpha+\psi(h)\to\alpha\), so \(f'(z_0)\) exists and equals \(\alpha\).
See [@SS03].