# Proving $f$ has at least one zero inside unit disk

Let $f$ be a non-constant and analytic on a neighborhood of closure of the unit disk such that $|f(z)|=\text{constant}$ for $|z|=1$. Prove $f$ has at least one zero inside unit disk.

I thought of using Rouche’s somehow. Using $f(z)-z$, and taking constant is less than $1$, I can actually conclude from Rouche’s theorem that the equation $f(z)-z=0$ have a fixed point inside the unit disk. I am stuck for other constants greater than equals to one and exactly zero.
I hope there should be a little trick I am missing here. It will be awesome to see if maximum principle can be applied to conclude the result.

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Hint: if $f$ has no zero inside the disk, consider $1/f$ and use the maximum modulus principle.

If you want to use Rouche’s Theorem, you can consider $g(z)=f(z)-f(w)$ for some $w\in D$ where $D$ is unit disc. Then on $\partial D$ we have $|g(z)-f(z)|=|f(w)|<|f(z)|$, since by maximum modulus principle the maximum of $|f(z)|$ can only happens on $\partial D$ and also we know that $|f(z)|$ is constant on $\partial D$. Then by Rouche’s Theorem we know that number of zeros of $f$ equals to it of $g$ on $D$. As $g$ at least has one zero $z=w$, $f$ at least has one zero.