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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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