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Fall 2019 - Problem 8

entire functions

A function $\func{f}{\C}{\C}$ is entire and has the property that $\abs{f\p{z}} = 1$ when $\abs{z} = 1$ . Prove that $f\p{z} = az^n$ for some integer $n \geq 0$ and some $a \in \C$ with $\abs{a} = 1$ .

Solution.

Since $f$ is not identically zero, its zeroes cannot accumulate, so in particular, $f$ can only have finitely many zeroes $a_1, \ldots, a_n \in \cl{\D}$ counting multiplicity. Since $f$ does not vanish on the boundary, we see that each $a_k \in \D$ . Thus, the Blaschke product

B\p{z} = \prod_{k=1}^n \frac{z - a_k}{1 - \conj{a_k}z}

is a holomorphic function such that $\abs{B\p{z}} = 1$ on $\partial\D$ with exactly the same zeroes as $f$ counting multiplicity. Thus, if $g = \frac{f}{B}$ , then $g$ is holomorphic and non-zero on $\D$ , so $\frac{1}{g}$ is also holomorphic on $\D$ . Observe that on $\partial\D$ ,

\abs{g\p{z}} = \frac{1}{\abs{g\p{z}}} = 1,

so by the maximum principle applied to both $g$ and $\frac{1}{g}$ , we get $\abs{g\p{z}} = 1$ on all of $\D$ . Thus, $g$ is constant on $\D$ , i.e., for any $z \in \D$ ,

g\p{z} = a \implies f\p{z} = a\prod_{k=1}^n \frac{z - a_k}{1 - \conj{a_k}z}

for some $a \in \C$ with $\abs{a} = 1$ . Thus, by uniqueness and connectedness, we have $f = B$ on their common domain of holomorphicity. If any $a_k \neq 0$ , then $f$ would have a pole at $1/\conj{a_k}$ , which is impossible since $f$ is entire. Thus, $f$ can only have zeroes at the origin, so $f\p{z} = az^n$ .