Show that if \(f\) is holomorphic on a connected region \(\Omega\) and \(f'\equiv 0\) on \(\Omega\), then \(f\) is constant on \(\Omega\).
Solution
Fix \(z_0\in\Omega\) and set \[ A=\{z\in\Omega:f(z)=f(z_0)\}. \] The set \(A\) is closed by continuity. It is also open: if \(z\in A\), choose a disk \(D(z,r)\Subset\Omega\). For any \(w\in D(z,r)\), integrate along the line segment from \(z\) to \(w\): \[ f(w)-f(z)=\int_z^w f'(\zeta)\,d\zeta=0. \] Hence \(D(z,r)\subset A\). Thus \(A\) is nonempty, open, and closed in the connected set \(\Omega\), so \(A=\Omega\). Therefore \(f\) is constant.