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$f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$

What I think is since $f$ do not have any zero for some bounded domain, I can define a branch of logarithm $(\log f)$ on that domain which will gives my desired result $f =e^{\log f}$. I don't know if I am doing it right? If this is right I don't know how do I argue $(\log f)$ is entire. Hint please.

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Just saying "a branch of logarithm" won't do it. In fact, since the range of $f$ will contain all nonzero complex numbers (see Picard's theorem) you can't choose a particular branch of the logarithm and have $\log f$ be entire.

Hint: $g'(z) = f'(z)/f(z)$.

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    $\begingroup$ If, after working with this hint, you still cannot get it, ask again...explaining your thoughts. $\endgroup$
    – GEdgar
    Dec 30 '12 at 20:52
  • $\begingroup$ @GEdgar, I can see $f= e^g$ is coming along with the hint, and $g^`$ being well defined and as a ($ g(z)= dlogf(z)$) as $f$ do not have any zero. I don't know though where I am going with that. Seems something going to happen with argument principle. I am lost. Help please. $\endgroup$
    – Deepak
    Dec 31 '12 at 1:08
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    $\begingroup$ Further hint: analytic functions have antiderivatives. $\endgroup$ Dec 31 '12 at 1:54
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    $\begingroup$ Do you know the theorem that an entire function (or more generally, an analytic function on a simply connected domain) has an antiderivative? If $g$ is an antiderivative of $f'(z)/f(z)$, can you show that $f/e^{g}$ is constant? $\endgroup$ Dec 31 '12 at 9:48
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    $\begingroup$ If $f/e^g = c$, then $f = e^{g+\log(c)}$. $\endgroup$ Dec 31 '12 at 20:22

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