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.

This is an exercise from a book called "theory of complex functions" I am trying to solve:

Let $G$ be a bounded region, $f,g$ continuous and zero-free in $\overline{G}$ and holomorphic in $G$. With $$|f(z)|=|g(z)| \ \ \ \ \ \forall z \in \partial G$$ It follows that there exists a $\lambda \in S^{1}$ such that $f(z)= \lambda g(z) \ \ \ \forall z \in \overline{G}$

I dont know how to even begin.

Does anybody see how to begin? Please, do tell.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Here's a plan. For all of this, note that as $f$ and $g$ do not vanish at any point of $\overline G$ we can make a lot of arguments using $f/g$ and its reciprocal.

We have a statement about absolute values on a boundary, which reminds us of the maximum modulus principle. Use this theorem to prove that $|f| = |g|$ on all of $\overline{G}$. What does the open mapping theorem tell us about a holomorphic function from $G$ to $S^1$?

share|improve this answer
    
We must require that "region" mean "connected" for this to work, I think. –  Dylan Moreland Dec 6 '11 at 14:30
    
Since$$|f(z)| = |g(z)|, z \in \partial G$$ with the maximum modulus principle:$$ \Rightarrow max|f|_{\overline{G}}=max|f| _{\partial G} = max|g| _{\partial G}= max|g| _{\overline{G}}$$$$\Rightarrow |f| _{\overline{G}}=|g| _{\overline{G}}$$ The open mapping theorem then tells us that because $G$ is open, the map of $S^{-1}$ is so, too. Thanks Dylan Moreland, is this what you meant so far? –  VVV Dec 6 '11 at 15:33
1  
@VVV I agree with your row of equalities, but I'm not sure that it helps: it isn't clear to me how you go from the maximums on $\overline G$ being equal to the absolute values being equal everywhere, which is what I think you mean in the last row. What I had in mind was that $|f/g| = 1$ on $\partial G$, and hence by the maximum modulus principle we have $|f/g| \leq 1$ on $\overline G$. Switching $f$ and $g$, we can get the reverse inequality, and hence equality $|f/g| = 1$ on all of $\overline G$. –  Dylan Moreland Dec 6 '11 at 15:49
    
Merci Dylan Moreland, but then how do you make a conclusion for $ \lambda$ in $S^{-1}$ and $|f|=\lambda |g|$ in $G$. Do you just put the $\lambda = 1 $ ? –  VVV Dec 6 '11 at 16:04
    
@VVV The fact that $f/g$ has constant modulus $1$ on all of $G$ means that it defines a holomorphic map $G \to \mathbf C$ which lands in $S^1$. You want to argue that this map is constant. –  Dylan Moreland Dec 6 '11 at 16:07

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.