Let $f$ be holomorphic on $\mathbb{D}$, continuous on $\overline{\mathbb{D}}$ s.t $|f(z)|=1$ whenever $|z|=1$. Prove that if $f(z)\neq 0$ for all $z\in\mathbb{D}$ then $f$ is constant on $\mathbb{D}$.
I already proved the following result:
Let $f$ be non-constant, holomorphic on $\mathbb{D}$, continuous on $\overline{\mathbb{D}}$ s.t $|f(z)|=1$ whenever $|z|=1$. Then there exists $z_0\in\mathbb{D}$ s.t $f(z_0)=0.$
I don't know whether this could help me solve the actual problem? Could someone give me any idea?
Let $$F(z)=\frac{1}{f(z)},\quad z\in\overline{\mathbb D},$$ then $F$ holomorphic on $\mathbb{D}$, continuous on $\overline{\mathbb{D}}$, and $|F(z)|=1$ whenever $|z|=1$. So $|F(z)|$ attains its minimum at some point $z_0\in\overline{\mathbb D}$.
If $|z_0|=1$, then $|F(z)|\equiv1$ on $\overline{\mathbb D}$ and $F\equiv C$;
If $|z_0|<1$, then $$|f(z_0)|=\max_{z\in\overline{\mathbb D}}|f(z)|.$$ maximum modulus principle implies $f$ is constant!