3
$\begingroup$

$f$ 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.

1 Answers 1

6

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

  • 1
    If, after working with this hint, you still cannot get it, ask again...explaining your thoughts.2012-12-30
  • 0
    @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.2012-12-31
  • 1
    Further hint: analytic functions have antiderivatives.2012-12-31
  • 0
    @RobertIsrael, Could you please tell me further what should I review to understand this whole thing, I am still dumb here.2012-12-31
  • 0
    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?2012-12-31
  • 0
    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.2012-12-31
  • 0
    If $f/e^g = c$, then $f = e^{g+\log(c)}$.2012-12-31