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Let $f$ be analytic and nowhere zero on $0<|z|<1$. Prove that $f(z)=z^n \exp(g(z))$ for some integer $n$ and $g$ analytic in $0<|z|<1$.

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Is that a fact? Is this a statement that you are trying to prove? Do you have any thoughts on the problem? – Jonas Meyer Dec 19 '12 at 2:20
Hint: let's define $g(z) = \ln f(z)$ and $f(z) = z^0 \cdot \exp(g(z))$. What's the problem with this “solution”? Why is $g(z)$ not well defined? How can we fix this problem? – Yury Dec 19 '12 at 3:48
One can define $ln(f(z))$ only in simply connected domain, so one can remove x axis from the domain and define $ln$. But I am not sure how to get on whole domain. If $f$ have a removable singularity or pole at $z=0$, I can see how to get result. What if it is an essential singularity? – Tits_Alternative Dec 22 '12 at 4:17

$\ln f$ is defined locally on $0<|z|<1$ as $\ln f = \ln |f| + i \arg f$ for some branch of $\arg f$. The issue is whether or not we have a single valued branch of $\ln f$ in the annulus $0<|z|<1$. We must check that analytic continuation along every (simple) closed path comes back to the original branch. Any such path is homotopic to either a constant path (nothing to worry about) or to a circle about the origin. Analytic continuation along such a circle yields the original branch plus $+2\pi i n$ for some $n\in\mathbb Z$, due to the aforementioned connection with argument. If $n\ne 0$ we are in trouble. To get out of trouble, apply the above not to $f$ itself, but to its product with an appropriate power of $z$.

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+1 The intuitive explanation works better for me than the technical details. – copper.hat Dec 24 '12 at 17:58

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