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Let $G$ be an open connected set and $f, g$ analytic functions on $G$. If $|f|\le |g|$ then there exists an analytic function $h$ such that $f(z)=h(z)g(z)$.

We know $|f/g|\le 1$ everywhere in $G$, and so $f/g$ is bounded near any singularities. (The singularities must be isolated, for otherwise they have a limit and the function is identically zero.) So we can extend $f/g$ to be analytic in all of $G$.

But where do I go from here? Presumably the connectedness of $G$ is important, but I don't see how it helps in manipulating $f$ and $g$ to make an $h$ appear

EDIT: As pointed out in the comments, I am already done. Take $h= f/g$.

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What's the problem? You already extended $f/g$ to be analytic in $G$. That's your $h$. –  Robert Israel Jan 16 '13 at 3:24
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Excuse my while I facepalm myself to death. Thank you –  Bey Jan 16 '13 at 3:26
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Bey: You might as well post an answer. –  Jonas Meyer Jan 16 '13 at 3:31

1 Answer 1

As has been pointed out, there's actually nothing left for me to do. For completeness, here's the solution:

Since $|f(z)|\le |g(z)|$ for all $z\in G$, we have $|f(z)/g(z)|\le 1$ everywhere in $G$. So $h(z):=f(z)/g(z)$ is bounded in a neighborhood of any possible singularities, meaning any singularity is removable. Hence, we can define $h$ to be analytic on all of $G$. Now we simply notice that $f(z)=h(z)g(z)$ and we're done.

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