If \(f(z)\) is analytic in \(|z| < 1\), continuous on \(|z| \leq 1\) and satisfies \(|f| = 1\) on \(|z| = 1\), show that \(f(z)\) is rational.
Hint: First consider the case that \(f(z)\) has no zero.
Solution
Because \(|f|=1\) on the unit circle and \(f\) is continuous on the closed disk, there is an annulus near \(|z|=1\) on which \(f\) has no zeros. Thus all zeros of \(f\) lie in a smaller closed disk, and hence there are only finitely many of them, say \(a_1,\dots,a_m\), repeated according to multiplicity.
For each \(a_j\) let \[ B_{a_j}(z)=\frac{z-a_j}{1-\overline{a_j}z}. \] Each \(B_{a_j}\) is rational, holomorphic on the unit disk, has its only zero at \(a_j\), and satisfies \(|B_{a_j}(z)|=1\) for \(|z|=1\). Set \[ B(z)=\prod_{j=1}^m B_{a_j}(z). \] Then \[ h(z)=\frac{f(z)}{B(z)} \] extends holomorphically and without zeros across the zeros of \(f\). Moreover \(|h|=1\) on \(|z|=1\).
By the maximum modulus principle, \(|h|\le1\) in the disk. Since \(h\) is zero-free, \(1/h\) is holomorphic, and the same argument gives \(|h|\ge1\). Hence \(|h|=1\) throughout the disk, so \(h\) is constant by the open mapping theorem. Therefore \[ f(z)=c\prod_{j=1}^m\frac{z-a_j}{1-\overline{a_j}z}, \qquad |c|=1, \] which is rational.