Let $f:\mathbb{C}\rightarrow\mathbb{C}$ entire function such that $\left|f(z)\right|=1$ when $|z|=1$. Show that $f(z)=Cz^{m}$ for some $m\in \mathbb{N}$ and $C\in\mathbb{C}$ with $|C|=1$.
My attempt: Since $f$ is entire we have that power series at $0$ $$f(z)=\sum_{n=0}^{\infty}c_{n}z^{n}$$ is convergent in all $\mathbb{C}$. I show that $|c_{n}|\leq 1$ for all $n\in\mathbb{N}$.
We have $f(z) = \sum_{n=0}^\infty c_n z^n$.
Let $g(z) = \overline{f(1/\overline{z})}=\sum_{n=0}^\infty \overline{c_n} z^{-n}$. It is analytic on $|z| > 0$.
Note that for $|z| = 1$ : $g(z) = \overline{f(z)}$, and since $|f(z)| = 1$ we have $g(z) f(z) = 1$.
By the identity theorem this means $fg = 1$.
Let $m$ be the order of vanishing of $f(z)$ at $z=0$. Then $g(z) = O(z^{-m})$. Therefore $\frac{f(z)}{z^m}$ is a bounded entire function, so by Liouville's theorem it is constant.
Qed $f(z) = A z^m$ for some $m \in \mathbb{N}$ and $|A|=1$.