Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 0 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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.