Suppose that $f:A\to \mathbb{C}$ analytic function on a convex region $A$, and f does not equal to zero in any point in $A$. Is it always true that there is such analytic function $g(z)$ $ f=e^g$ on $A$?
Exponential function
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complex-analysis
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0Yes, I solved it thanks, I am posting answer now – 2012-10-31
2 Answers
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Take a function $f'(z)/f(z)$, by given assumptions it is analytic on convex region, so it has analytic primitive $F(z)$. Let $h(z)=f(z)^{-F(z)}$. Then we find that $h'=0$ on $A$. So $h$ equals to some nonzero constant, $ fe^{-F}=c$ Let $a\in \mathbb{C}$ be such that $e^a=c$, so we got that $f=e^{F+a}$ and $F+a$ is analytic function.
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0Very good. Observe that $f'/f$ formally looks like the derivative of $\log{f}$ (it's called the [logarithmic derivative](http://en.wikipedia.org/wiki/Logarithmic_derivative)), so there's no surprise that its primitive $F$ does what it does... – 2012-10-31
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Note: this answer an earlier version of the question.
No. If $A$ is multiply connected then there is no such a $g$.
EDIT: after the edit of the question, then yes there must exist $g$ such that $f=e^g$.