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Spring 2016 - Problem 9

entire functions

Let $\func{f}{\C}{\C}$ be entire and assume that $\abs{f\p{z}} = 1$ when $\abs{z} = 1$ . Show that $f\p{z} = Cz^m$ , for some integer $m \geq 0$ and $C \in \C$ with $\abs{C} = 1$ .

Solution.

Since $f$ is non-zero on $\partial\D$ , in particular $f$ is not identically zero. Hence, we may enumerate its zeroes $\set{a_n}_n$ . Let

B\p{z} = \prod_{\abs{a_n} < 1} \frac{z - a_n}{1 - \conj{a_n}z},

which is a finite product, since $f$ can only have finitely many zeroes on the compact set $\cl{\D}$ . Since $\abs{B\p{z}} = 1$ when $\abs{z} = 1$ , we see that both $\abs{f/B} = 1$ and $\abs{B/f} = 1$ on $\partial\D$ . By the maximum principle, we see that $f\p{z} = CB\p{z}$ for some constant $C \in \C$ in $\D$ . On the unit circle, we have

\abs{f\p{z}} = \abs{C} \abs{B\p{z}} \implies \abs{C} = 1.

By uniqueness, we see that $f\p{z} = CB\p{z}$ on all of $\C$ . If any of the $a_n$ are non-zero, however, we see that $f$ would have a pole at $\frac{1}{\conj{a_n}}$ , which is impossible as $f$ is entire. Thus, the only zeroes of $f$ are at the origin, so $f\p{z} = Cz^m$ for some integer $m \geq 0$ .