TITLE: $f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$
QUESTION [8 upvotes]: $f$ 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.

REPLY [10 votes]: 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: $g'(z) = f'(z)/f(z)$.