In a solution to a problem, I read that, if $f(z)$ is entire, $f(z)\neq0$ and the domain of definition of $f(z)$ is simply connected, then it is possible to choose a branch of log $f(z)$ that is analytic in the entire plane.

I was a bit surprised by this. My understanding is that, given at point $z_0\neq0$, you can always choose a branch so that log $z$ is analytic at that point. But once you have chosen a branch, that branch will not be analytic in the branch cut. To me this implies, that you can never choose $one$ branch of the logarithm, such that log $f(z)$ is analytic in the whole plane.

Grateful if you can sort this out for me. Where do I go wrong? Am I misinterpreting what is said in the solution?

  • $\begingroup$ If $f$ is entire then its domain is $\mathbb{C},$ which is simply connected - no need to state that assumption. $\endgroup$ – zhw. May 20 '15 at 19:25
  • $\begingroup$ What was the original problem and its source? $\endgroup$ – whacka May 20 '15 at 19:43

If $f$ is holomorphic on a simply connected domain $U$ and $0\not\in f(U)$ then there exists another function $g$ also holomorphic on $U$ such that $f(z)=\exp g(z)$. In fact, observe that

$$f(z)=e^{g(z)}\implies f'(z)=g'(z)e^{g(z)}=g'(z)f(z)\implies g'(z)=\frac{f'(z)}{f(z)}$$

which inspires $g(z):=\displaystyle \int_w^z\frac{f'(\xi)}{f(\xi)}d\xi$ (this is well-defined since $U$ is simply connected).

It is not however true that $\log f(z)$ can be defined for a continuous single-valued branch of $\log$, for instance consider $f(z)=e^z$ and $U=\Bbb C$ - then $f(U)=\Bbb C\setminus\{0\}$ is not simply connected.

  • $\begingroup$ Thanks for your answer, but I don't quite understand the difference. The first part of your answer states that it's possible to define a function $f(z)=e^{g(z)}$ such that $g(z)$ is analytic in the whole complex plane. Isn't that the same as saying that $g(z) =$ Log $f(z)$ is in fact entire? But then below you state that log $f(z)$ can't be defined for a single-valued branch for log... I'm confused. Please help me understand the difference here! $\endgroup$ – Jarvi79 May 21 '15 at 8:34
  • $\begingroup$ @Jarvi79 Did you read my $f(z)=e^z$ example at the end? Clearly there is a holomorphic $g$ for which $f(z)=e^{g(z)}$ (it's just $g(z)=z$). However there is no branch of the $\log $ function for which $\log e^z$ is holomorphic, and in particular $\log e^z$ is not the same as $z$ (I never said anything about $g(z)$ being $\log f(z)$ by the way). Indeed, whatever branch you pick, the imaginary part of $\log e^z$ will jump discontinuously as $e^z$ crosses that branch, right? $\endgroup$ – whacka May 22 '15 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.