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Does the existence of a holomorphic square root for the identity function in a region $\Omega$ in $\mathbb C$ imply the existence of a holomorphic logarithm for the same function? I have no idea how to prove this.

Kavi Rama Murthy
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  • I believe this is false but I have no idea how to construct a counter-example. – Kavi Rama Murthy Jan 28 '18 at 11:42
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    Well, it is ultimately true that the regions where there is a holomorphic section of $z^2$ are exactly the regions where there is a holomorphic section of $\exp$. –  Jan 28 '18 at 11:45
  • Can you characterize the regions which admit a square root? What about the regions that admit a log? – Aaron Jan 28 '18 at 11:45
  • Possible duplicate of: https://math.stackexchange.com/questions/3736/a-criterion-for-the-existence-of-a-holomorphic-logarithm-of-a-holomorphic-functi – preferred_anon Jan 28 '18 at 11:48
  • A clarification: what I wanted to assume was that the identity function had a square root in $\Omega$. In other words, there exists a holomorphic function f on $\Omega$ with ${f(z)}^{2}=z$ for all $z \in\Omega$. With no further assumption can we say there exists a holomorphic function g on $\Omega$ with $e^{g(z)}=z$ for all $z \in \Omega$. – Kavi Rama Murthy Jan 28 '18 at 13:25
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    The question as stated is unclear - you're assuming what has a square root? Instead of just clarifying in a comment you should edit the question!!! – David C. Ullrich Jan 28 '18 at 17:05
  • @Daniel Littlewood , it doesn't appear to me that this is a duplicate. I have edited the question to make the hypothesis more explicit. – Kavi Rama Murthy Jan 29 '18 at 08:14
  • @KaviRamaMurthy This is progress. Now edit again to clarify the conclusion and things are great. – David C. Ullrich Jan 30 '18 at 16:32

1 Answers1

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Answer to the question as clarified in a comment: If $z$ has a holomorphic square root in $G$ does $z$ have a holomorphic logarithm?

The answer is yes. We use the following intuitively clear statement:

Lemma. Suppose $G\subset\Bbb C$ is open, $0\notin G$, and some closed curve in $G$ has non-zero index (winding number) about the origin. Then some closed curve in $G$ has index $1$ about the origin.

For an informal proof see here. Assuming that, suppose $g^2=z$ (which of course implies $0\notin G$). Then $2gg'=1$, so $$2\frac{g'}g= 2\frac{g'g}{g^2}=\frac 1z.$$

Then for every closed curve $C$ we have $$\frac1{2\pi i}\int_C\frac 1z =2\frac{1}{2\pi i}\int_C\frac{g'}g.$$Since $\frac{1}{2\pi i}\int_C\frac{g'}g$ is just the index of $g\circ C$ about the origin it is an integer; hence $\frac1{2\pi i}\int_C\frac 1z$ is an even integer for every $C$. The Lemma now implies that $\frac1{2\pi i}\int_C\frac 1z=0$ for every $C$, so that $1/z$ has a primitive.

David C. Ullrich
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