# Ngô Quốc Anh

## July 18, 2009

### Schwarz Reflection Principle and several applications, 2

I want to continue the topic “Schwarz Reflection Principle and several applications”. Today we discuss the following.

Question. If entire function $g$ satisfies $|g(z)|=1$ whenever $|z|=1$, then there exist a non-negative integer number $n$ and a constant $c$ satisfying $|c| = 1$ such that $g(z) = cz^n$.

Solution. Suppose $g$ has a zero at the origin of order $n$ for some $n\ge 0$ and the other zeros of $g$ in the unit disk are listed as $a_1,a_2,\dots,a_m$, repeating as necessary to account for multiplicities. The set of zeros in the unit disk must be finite since that set cannot have any limit points.

Let

$\displaystyle B(z) = z^n\prod_{j = 1}^m\frac {z - a_j}{1 - \overline{a_j}z}$.

Note that $|B(z)| = 1$ for all $|z| = 1$. Let $f(z) = \frac {g(z)}{B(z)}$. Then $f$ is an entire function with no zeros in the unit disk and with the property that $|f(z)| = 1$ for $|z| = 1$.

By the maximum modulus principle $|f(z)|$ and $\frac {1}{|f(z)|}$ must both achieve their maximum values for the unit disk on the boundary, from which we conclude that $|f(z)| = 1$ for all $|z|\le 1$. But then, that means that $f(z) = c$ must be constant, with $|c| = 1$.

So we have that

$\displaystyle g(z) = cB(z) = cz^n\prod_{j = 1}^m\frac {z - a_j}{1 - \overline{a_j}z}$.

But that expression has poles whenever $z = \frac1{\overline{a_j}}$. The only way we could have $g$ be an entire function is for there to be no such $a_j$. Hence we conclude that $g(z) = cz^n$ for some nonnegative $n$.

Here is an even further simplification.

Dividing by some finite power of $z$ (which does not change the property), we can assume that $g(0)\ne 0$. Now consider

$\displaystyle \frac1{g(1/z)} \quad \text{ and } \quad \overline{g(\bar z)}$.

They are meromorphic on $\mathbb C$ and coincide on the unit circle. Thus they coincide everywhere. Hence

$\displaystyle g(z)\to \frac1{g(0)} \quad \text{ as } \quad z\to\infty$,

so $g$ is bounded and, thereby, constant.

I will provide another proof mainly based on the Schwarz Reflection Principle. However, to this purpose, an extension of Morera’s Theorem for toy contours should be introduced firstly.