If $f$ is an entire non-constant function satisfying $|f(z)|=1 \ \forall z\in A=\{z: |z| = 1 \}$ then there exists $z\in \mathbb D (0,1)$ such that $f(z)=0$.
I'm trying to prove this following this hint, but I don't really know how to use it.
Suppose $f$ doesn't vanishes in $\mathbb D (0,1)$. Then $1/f$ is well defined and it is a composition of holomorphic functions and thus holomorphic in $\mathbb D$, satisfying $1/|f(z)|=1$ in $A$. How can we use the Maximum principle here?
The function $\left\lvert\frac1f\right\rvert$ has a maximum on $\overline{\mathbb D}$ (since $\overline{\mathbb D}$ is compact). If that maximum is greater than $1$, it is loccated at $\mathbb D$ and therefore, by the maximum principle, $\frac1f$ is constant, which is impossible, since $\bigl\lvert f(z)\bigr\rvert=1$ when $\lvert z\rvert=1$. So, a contradiction is reached, and $\frac1f$ doesn't exist. So, $f$ has a zero.
And if the maximum is located at the boundary of $\mathbb D$, that $\lvert f\rvert=1$, and, again by the maximum principle, $f$ is constant.