Assume \(f(z)\) is analytic in a region \(D\) and \(\Gamma\) is a simple closed rectifiable curve in \(D\) whose interior \(\Omega\) is contained in \(D\). Prove that if \(f(z)\) is real-valued for all \(z \in \Gamma\), then \(f(z)\) is constant on \(D\).
Solution
Write \[ f=u+iv. \] Since \(f\) is holomorphic, \(v\) is harmonic on \(D\). By hypothesis, \[ v=0\quad\text{on }\Gamma=\partial\Omega. \] Because \(\Gamma\) is a simple closed curve and \(\Omega\) is its bounded interior, \(\overline\Omega=\Omega\cup\Gamma\subset D\).
1. The harmonic maximum principle applied to \(v\) on \(\Omega\) gives\[ v\le0 \]throughout \(\Omega\), because the boundary values are zero. Applying it to \(-v\) gives\[ -v\le0. \]Hence\[ v\equiv0\quad\text{on }\Omega. \]
2. Thus \(f(\Omega)\subset\mathbb R\). If \(f\) were nonconstant on the connected open set \(\Omega\), the open mapping theorem would imply that \(f(\Omega)\) is open in \(\mathbb C\), impossible for a subset of \(\mathbb R\). Therefore \(f\) is constant on \(\Omega\).
3. Since \(D\) is connected and \(f-c\) vanishes on the nonempty open subset \(\Omega\), the identity theorem gives\[ f\equiv c\quad\text{on }D. \]