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Suppose f is meromorphic in a neighborhood of the closed unit disk , that it does not have zeroes nor poles in the open unit disk, and that $|f(z)|=1$ for $|z|=1$. Find the most general such function. Let's denote D = open unit disk

Well, since f has no poles in D, it's holomorphic there, thus by the maximum modulus principle $ |f(z)| < 1$ for $|z|<1$. If f does not have a zero , then we can use the minimum modulus principle, so f attains it's minimum in $\partial D $ thus $f(z)=1 \forall z \in D$ , by analytic continuation $f(z)=1 \forall$ in where f is defined $ I'm not sure if my solution it's correct :S. I never used the fact that it's analytic in a neighborhood of the closure. I'm missing something?

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up vote 1 down vote accepted

Actually the conclusion from the minimum modulus principle is $|f(z)| = 1$, not $f(z) = 1$. And the And then you can use the Open Mapping Theorem to say that $f$ is constant.

You did use the fact that $f$ is continuous on the closed unit disk in applying the maximum and minimum modulus principles.

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You are completely right, I use the hypothesis that my domain is open , to assert that $f(D)$ is open but contained in $S^1$ Thanks :D! – Daniel Oct 15 '12 at 20:12

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