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
0
$\begingroup$
complex-analysis
-
1Robert Israel's answer [here](http://math.stackexchange.com/q/143609/16627) gives the general strategy. I suggest that you try to work this out on your own. – 2012-10-29
-
0Did 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. – 2012-10-31
-
0Yes, I solved it thanks, I am posting answer now – 2012-10-31
2 Answers
1
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.
-
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
2
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 $f=e^g$.