Let \(f\) be analytic on a bounded domain \(D\), and assume also that \(f\) that is continuous and nowhere zero on the closure \(\bar{D}\).
Show that if \(|f(z)|=M\) (a constant) for \(z\) on the boundary of \(D\), then \(f(z)=e^{i \theta} M\) for \(z\) in \(D\), where \(\theta\) is a real constant.
Solution
By the maximum modulus principle, \(\abs{f} \leq M\) in \(\bar{D}\). Since \(f\) has no zeros in \(\bar{D}\), \(g\da 1/f\) is holomorphic on \(D\) and continuous on \(\bar{D}\). So the maximum modulus principle applies to \(g\), and \(\inverseof{M} \geq \abs{g} = 1/\abs{f}\), so \(\abs{f} \geq M\). Combining these, \(\abs{f(z)} = M\) for every \(z\in D\). Then \(f(D)\) lies in the circle \(\abs{w}=M\), which contains no nonempty open set, so by the open mapping theorem \(f\) is constant on the domain \(D\): \(f(z) = \lambda M\) where \(\lambda\) is some constant with \(\abs{\lambda}=1\). This is on the unit circle, so \(\lambda = e^{i\theta}\) for some fixed angle \(\theta\).