# $G$ a bounded region, $f,g$ continuous and zero free follows $f(z)=\lambda g(z)$ for some $\lambda \in \partial G$

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.

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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 $S^1$?
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
@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