Determine all entire functions $f$ with the property that if $|z|=1$, then $|f(z)|=1$.

Question

My question is how to solve: Determine all entire functions $f$ with the property that if $|z|=1$, then $|f(z)|=1$.

I was thinking that I could solve this by first show that $f$ has to be a polynomial. Then I can use the formula \begin{align*} f(z)=z^{n}(a_{n}+a_{n-1}/z+...+a_{1}/z^{n-1+a_{0}/z^{n}}) \end{align*} where $a_{n}\neq 0$. And then somehow show that \begin{align*} f(z)=z^{n}a_{n}, \end{align*} with $|a_{n}|=1$. (I think that this is the answer.)

Could someone help me? What are the steps I should to?

Some theorems that my book covers and I think will be needed are Liouville's theorem, Maximum principle, fundamental theorem of algebra. Thanks!

Answer 1

Disclaimer: Wrong proof, see commments. I will leave it up for discussion purposes and edit it later. Think this is more about the maximum modulus principle. By that we have that $|f(z)|\leq 1$ within the unit disk. Also, similarly, by the minimum modulus principle, we have that the function either has a zero or we have $|f(z)|\geq1$.
So lets assume the function has no zero. Then we can conclude that $|f(z)|=1$ for all $z \in D$. Since $|f|$ is constant, so is $f$. So any function with $f(z)=w$, $|w|=1$ is a suitable candidate.
Lets assume the function has a zero of order n, lets say in $z_0$. Then we have $$ f(z)=(z-z_0)^nh(z) $$ Since $h(z)$ is zero-free within $D$, we can apply the min/max principle again. In absolute values, this will lead to: $$ \frac{1}{|z-z_0|^n} \leq|h(z)| \leq \frac{1}{|z-z_0|^n} $$
But that means that our function $f(z)$ statisfies the estimate for $z \in D$: $$ 1 \leq |f(z)|\leq 1 $$ As above, $f(z)$ has to be constant, but it has a zero. A contradiction.

Answer 2

Hint: For $a\in \mathbb D,$ the open unit disc, define

$$g_a(z) = \frac{a-z}{1-\bar a z}.$$

Then $g_a$ is holomorphic on a neighborhood of the closed unit disc, and $|g|=1$ on $\partial \mathbb D.$ Any finite product of such functions is a candidate here.