Complex Analysis – Nonconstant Holomorphic Function in Unit Disk

Question

$f$ be a non-constant holomorphic in unit disk such that $f(0)=1$. Then it is necessary that

1. there are infinitely many points inside unit disk such that $|f(z)|=1$

2. $f$ is bounded.

3. there are at most finitely many points inside unit disk such that $|f(z)|=1$

4. $f$ is rational function.

Counter examples for $2,4$ are ${1\over z-1},e^z$

If there are infinitely many points in the open disk such that $|f(z)|=1$ then that set is being bounded. Since there are infinite points so it must have a limit point in it and $f$ will be bounded. And hence by Liouvilles Theorem, it is constant. This is a contradiction

So $3$ is the correct statement.
Thank you for confirming.

Answer 1

Hint: For $1,3$ use the Open Mapping Theorem with the open unit disk to prove that there are infinitely many points.