Let $f:D(0,1)\rightarrow \mathbb{C}$ be analytic such that $f(0)=1$. Prove that there are an infinite number of distinct points $z_{n}$ in $D(0,1)$ such that $|f(z_{n})|=1$
Remark: $D(0,1)$ is the open unit circle
My thought on this matter is to try and prove by contradiction (with the maximum modulus principle).
Assume that there are a finite number of points $z_{n}$ in such that $|f(z_{n})|=1$
Obviously $z=0$ is one of them, we assume WLOG that $z_{1}=0$
Define $R=min\left \{ |z_{2}|,...,|z_{n}| \right \}$
Let us pick $0<r<R$
By the assumption, $\left |f(z) \right |<1$ or $\left |f(z) \right |>1$ for $0<|z|<r$.
If $\left |f(z) \right |<1$ for $0<|z|<r$ then we get a contradiction to the maximum modulus principle since $f(0)=1$.
If $\left |f(z) \right |>1$ for $0<|z|<r$ then by the minimum principle we again get a contradiction
Is my solution correct?