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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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    See also http://math.stackexchange.com/questions/1565376/if-f-is-a-non-constant-analytic-function-on-b-such-that-f-is-a-constant2015-12-11

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