This question is in light of a previous question
- There are infinitely many points inside unit disk such that $|f(z)|=1$
- $f$ is bounded.
- There are at most finitely many points inside unit disk such that $|f(z)|=1$
- $f$ 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?