Assume \(f\) is an entire function such that \(|f(z)|=1\) on \(|z|=1\). Prove that \(f(z)=e^{i \theta} z^{n}\), where \(\theta\) is a real number and \(n\) a non-negative integer.
Suggestion: First use the maximum and minimum modulus theorem to show \[ f(z)=e^{i \theta} \prod_{k=1}^{n} \frac{z-z_{k}}{1-\bar{z_{k}} z} \] if \(f\) has zeros.
Solution
First show the hint: assume \(f\) has nonzero zeros. Write \(Z(f) \da \inverseof{f}(0)\) for the set of zeros in \(\bar{\DD}\).
If we assume \(f\) is continuous on \(\DD\), then \(\size Z(f) < \infty\)
Suppose \(\size Z(f) = \infty\), then by compactness of \(\bar{\DD}\) there is a limit point \(z_0\). If \(z_0 \in \DD\), then there is a sequence \(\ts{z_k}\to z_0\) with \(f(z_k) = 0\) for every \(k\), so \(f\) is zero on a set \(S\da \ts{z_k}_{k\geq 1} \union \ts{z_0}\) with an accumulation point and this forces \(f\equiv 0\) on \(\bar{\DD}\) by the identity principle, contradicting \(\abs{f} = 1\) on \(\bd \DD\)>
Otherwise, if \(z_0\in \bd \DD\), using continuity of \(f\) we have \(f(z_k) = 0\) for all \(k\) and \(z_k\to z_0\) so \(f(z_0) = 0\), again contradiicting \(\abs{f} = 1\) on \(\bd \DD\).
So write \(Z(f) = \ts{z_1,\cdots, z_m}\) and define \[ g(z) \da \prod_{1\leq k \leq m} {z-z_k \over 1 - \bar{z_k} z}, \quad h(z) \da {f(z) \over g(z)} .\]
\(h(z) \equiv 1\) is constant on \(\bar{\DD}\), so that \(f = \lambda g\) for some \(\lambda \in S^1\), i.e. \(\lambda = e^{i\theta}\) for some \(\theta\).
Note that \(h\) cancels all zeros of \(f\), so \(h\) is nonzero and holomorphic on \(\bar{\DD}\). Moreover \(\abs{g(z)} \leq 1\) on \(\bar{\DD}\) since these are well-known to be in \(\Aut(\DD)\). It’s also well-known that \(\abs{g(z)} = 1\) on \(\bd \DD\). Thus \(\abs{h(z)} = 1\) and \(\abs{1\over h(z)} =1\) on \(\bd \DD\), and by the maximum modulus principle, \[ \abs{h(z)} \leq 1 \quad\text{ and }\quad \abs{1\over h(z)}\leq 1 \quad \text{ on } \DD ,\] forcing \(\abs{h(z)}\equiv 1\) and thus \(h(z) = e^{i\theta}\) for some \(\theta\).
So we now have \[ f(z) = e^{i\theta} \prod_{1\leq k\leq m} {z-z_k \over 1 -\bar{z_k} z} ,\] which has poles at points \(z\) for which \(\bar{z_k}z=1\) for some \(z_k\in Z(f)\). However, since we assumed \(f\) was entire, it can have no such poles, which forces \(z_k = 0\) for all \(k\). But then \[ f(z) = e^{i\theta}\prod_{1\leq k \leq m}{z- 0 \over 1 - 0\cdot z} = e^{i\theta}z^m .\]