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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 $A$?

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Robert Israel's answer here gives the general strategy. I suggest that you try to work this out on your own. – commenter Oct 29 '12 at 11:35
ok, thank you very much – StudentMath Oct 29 '12 at 11:39
Did you manage to find the answer? If so, it would be great if you posted an answer containing the argument. If not, please ask what's still troubling you. – commenter Oct 31 '12 at 6:31
Yes, I solved it thanks, I am posting answer now – StudentMath Oct 31 '12 at 11:17

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 $f=e^g$.

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oh sorry, I made mistake $A$ is convex region, I will correct it now – StudentMath Oct 29 '12 at 11:20
up vote 1 down vote accepted

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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Very good. Observe that $f'/f$ formally looks like the derivative of $\log{f}$ (it's called the logarithmic derivative), so there's no surprise that its primitive $F$ does what it does... – commenter Oct 31 '12 at 11:26

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