8
$\begingroup$

$f$ 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.

$\endgroup$

1 Answer 1

10
$\begingroup$

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

$\endgroup$
10
  • 1
    $\begingroup$ If, after working with this hint, you still cannot get it, ask again...explaining your thoughts. $\endgroup$
    – GEdgar
    Commented Dec 30, 2012 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
    Commented Dec 31, 2012 at 1:08
  • 1
    $\begingroup$ Further hint: analytic functions have antiderivatives. $\endgroup$ Commented Dec 31, 2012 at 1:54
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Dec 31, 2012 at 9:48
  • 1
    $\begingroup$ If $f/e^g = c$, then $f = e^{g+\log(c)}$. $\endgroup$ Commented Dec 31, 2012 at 20:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .