Let $f$ be a non-constant analytic function in $D=1<|z|<2$ with $|f|\equiv5$ on the boundary (assume $f$ is continuous on $\overline D$). Show that $f$ has at least two zeros.
My opinion: Based on maximum module principle, we have $|f(z)|\leq 5$ for any $z \in D$. Suppose $f$ doesn't have zero in that region $D$, then we have $1/|f(z)|=1/5$ on the boundary and $1/|f(z)|<1/5$ (based on maximum module principle theorem and the fact that $f(z)$ is non-constant). However, it contracts to $|f(z)|\leq 5$, which implies that $|f(z)|\geq 1/5$ for any $z \in D$. That's why $f(z)$ has to have at least one zero in $D$. But how can I prove $f$ has at least two zeroes?