Holomorphic square root implies logarithm?

Question

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.

Answer 1

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.