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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.

Michael Hardy
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Deepak
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1 Answers1

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

Robert Israel
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    If, after working with this hint, you still cannot get it, ask again...explaining your thoughts. – GEdgar Dec 30 '12 at 20:52
  • @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. – Deepak Dec 31 '12 at 01:08
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    Further hint: analytic functions have antiderivatives. – Robert Israel Dec 31 '12 at 01:54
  • @RobertIsrael, Could you please tell me further what should I review to understand this whole thing, I am still dumb here. – Deepak Dec 31 '12 at 06:18
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    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? – Robert Israel Dec 31 '12 at 09:48
  • I did not know that theorem. May be I did not use that for any problem, or It could be the case I just forgot that. But the problem seems obvious with this hint. But shall I choose the path of integration is from $e^1$ to $z$ so that the constant is exactly 1? Can I do that? or it does not matter to get any constant? I can see the ratio $f/e^g$ is constant whatever path of integration I choose. – Deepak Dec 31 '12 at 15:46
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    If $f/e^g = c$, then $f = e^{g+\log(c)}$. – Robert Israel Dec 31 '12 at 20:22
  • May I ask for clarification, can't we define a single valued analytic branch of $\log f(z)$ for $f(z)\neq 0$ in a simply connected region? In Ahlfors, after General Theorem of Cauchy (Thm 15, Corollary 2), he defines in the same way suggested $e^{F(z) - F(z_0) + \log(f(z_0))} = f(z)$, and $F(z) - F(z_0) + \log(f(z_0)) = \log f(z)$. – Jo' Jul 03 '19 at 12:25
  • Yes, you can define a single valued branch of $\log f(z)$ in a simply connected region. But this is not done by taking a particular branch of $\log$ and composing that with $f$. – Robert Israel Jul 03 '19 at 14:09
  • Ah I see, thanks. – Jo' Jul 04 '19 at 00:49