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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.

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closed as off-topic by José Carlos Santos, Parcly Taxel, Shailesh, user99914, TheSimpliFire Feb 4 '18 at 10:07

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  • $\begingroup$ I believe this is false but I have no idea how to construct a counter-example. $\endgroup$ – Kavi Rama Murthy Jan 28 '18 at 11:42
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    $\begingroup$ 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 $\exp$. $\endgroup$ – user228113 Jan 28 '18 at 11:45
  • $\begingroup$ Can you characterize the regions which admit a square root? What about the regions that admit a log? $\endgroup$ – Aaron Jan 28 '18 at 11:45
  • $\begingroup$ Possible duplicate of: math.stackexchange.com/questions/3736/… $\endgroup$ – preferred_anon Jan 28 '18 at 11:48
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    $\begingroup$ 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!!! $\endgroup$ – David C. Ullrich Jan 28 '18 at 17:05
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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.

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  • $\begingroup$ Thank you. The negative votes by others were not justified. $\endgroup$ – Kavi Rama Murthy Jan 28 '18 at 22:31
  • $\begingroup$ @KaviRamaMurthy Could be those downvotes were because the question was unclear, not because it was trivial or stupid. You still haven't edited the question to clarify the hypotheses.... $\endgroup$ – David C. Ullrich Jan 28 '18 at 22:49

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