Liouville's theorem
Proof. Suppose \(\abs {f} \leq M\) on \(\C \). Fix \(z_0\). For every \(R > 0\), Theorem 3.1 with \(n = 1\) gives \(\abs {f'(z_0)} \leq M / R\). Letting \(R \to \infty \) gives \(f'(z_0) = 0\). So \(f' \equiv 0\) on the connected set \(\C \), and \(f\) is constant.