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## Entire function

Prove that if $f :\mathbb{C} \to \mathbb{C}$ is an entire function and $|z|=1 \Rightarrow |f(z)|=1$ then $f$ is constant or there exists $c \in \mathbb{C}$ with $|c|=1$ and $n$ some positive integer such that $f(z)=cz^n$.

Solution: $f$ cannot pe identically 0, therefore, the set of zeros of $f$ is a discrete set ( formed by isolated points ). Define $g(z)=\displaystyle \frac{1}{\overline{f(\frac{1}{\bar{z}}})}$ on the set $\Omega$ where $\frac{1}{\bar{z}}$ is never a zero for $f$. Check by definition that $g$ is also holomorphic on $\Omega$. Therefore, the equality $g(z)=f(z),\ |z|=1$ implies $f(z)=g(z)$ on $\Omega$.

Suppose $f$ has a zero $z_0 \neq 0$. Then $f$ is bounded in a punctured disk around $\frac{1}{\bar{z_0}}$, which should be a pole for $g$. But in a sufficiently small punctured disk around $\frac{1}{\bar{z_0}}$ $f=g$, and thus $g$ is bounded in a neighbourhood of this point, and the singularity is not a pole, but is removable. Contradiction.

Therefore, the only point where $f$ can be 0 is at $z=0$. Suppose now that $f(z)=z^k h(z)$, where $h(0)\neq 0$. Then $h$ satisfies the same conditions as $f$, but is never $0$. Defining an analogue $g$-function for $h$ also, we see that $h(\mathbb{C})\subset h(\mathbb{D})$ which is bounded. By Liouville’s Theorem, $h$ is constant. Therefore $f(z)=c z^k, \ k \geq 0, \ |c|=1$.