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 $0<|z|<1$.
Non zero analytic functions on annulus
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complex-analysis
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2Is that a fact? Is this a statement that you are trying to prove? Do you have any thoughts on the problem? – 2012-12-19
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0Hint: 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? – 2012-12-19
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1One 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? – 2012-12-22