# Nonconstant holomorphic function f in unit disk such that f(0)=1

This question is in light of a previous question

$f$ be a nonconstant 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.

2 is wrong by Liouville's theorem.

The author has used $f(z)=e^z$, (holomorphic and $f(0)=1$) to show that option 4 is wrong. Isn't $|f(z)|=1$ only for $z=0$ in this case? That means there are only finite (more specifically one) point inside unit disc such that $|f(z)|=1$.

But someone has proved that 1 is the right option.

Where am I going wrong?

With $f(z) = e^z$, $|f(z)| = e^{\operatorname{Re}(z)}$. Thus $|f(z)| = 1$ if and only if $\operatorname{Re}(z) = 0$, i.e., $z$ is purely imaginary. There are infinitely many imaginary numbers in the unit disk.