Suppose \(f\) is continuous and non-zero on \(\overline{\DD}\) and holomorphic on \(\DD\) such that \(|f(z)|=1\) for all \(|z|=1\). Show that \(f\) is constant.
Solution
Since \(f\) has no zeros on \(\overline{\mathbb D}\), the function \(1/f\) is holomorphic on \(\mathbb D\) and continuous on \(\overline{\mathbb D}\). On the unit circle, \[ |f|=1, \qquad |1/f|=1. \] The maximum modulus principle applied to \(f\) gives \(|f(z)|\le1\) in \(\mathbb D\), while applied to \(1/f\) it gives \(|f(z)|\ge1\). Hence \[ |f(z)|=1 \qquad(z\in\mathbb D). \] A nonconstant holomorphic function is an open map, but the unit circle has empty interior. Therefore \(f\) must be constant.