Holomorphic square root implies logarithm? [closed]

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.

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

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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
• 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 $\exp$. – user228113 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: math.stackexchange.com/questions/3736/… – preferred_anon Jan 28 '18 at 11:48
• 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

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