Does a branch of a square root determine a branch of a logarithm? Suppose I have a branch of the logarithm, that is, a continuous function $L(z)$ on some region $\Omega$ such that $e^{L(z)} = z$. We see that this defines a branch for the square root function on $\Omega$, via $\sqrt{z} = \exp(1/2 L(z))$, since 
$$(\exp(1/2 L(z))^2 = \exp(L(z)) = z$$
I am wondering if a sort of converse of this holds. Suppose on the other hand, we have a branch for the square root, i.e. some continuous function $R(z)$ on $\Omega$ such that $R(z)^2 = z$. Is there some way to get a branch of the logarithm from $R(z)$? If so, does this generalize (i.e. what branches for multi-valued functions will determine a branch of the logarithm)?
 A: While every cut for the square-root function is admissable as a cut for the logarithm in $\mathbb{C}\backslash\{0\}$, the complex plane without the origin, there are only two "layers" of the square-root function and countably many for the logarithm.  So it cannot be said that a branch of the square-root function "determines" a branch of the logarithm, unless you are willing to simply impose some choice of branches ad hoc.
A: Suppose $\Omega=\{z\in\mathbb{C}:z\nleq 0\}$. For $z\in\Omega$, let $\mathrm{Arg}(z)$ indicate the unique $\theta\in(-\pi,\pi)$ such that $z=|z|e^{i\theta}$. Define $R(z)=|z|^{1/2}e^{\frac{1}{2}\mathrm{Arg}(z)}$. This is a branch of the square root.
For each $k\in\mathbb{Z}$, $z\in\Omega$, let $L_k(z)=\log|z|+i(\mathrm{Arg}(z) + 4\pi k)$. Each $L_k$ is a distinct branch of the logarithm, but each one should satisfy $R(z)=\mathrm{exp}(\frac{1}{2}L_k(z))$.
A: A region $\Omega\subset{\mathbb C}$ admits a continuous branch $R$ of $z\mapsto\sqrt{z}\ $ iff $\ \Omega$ contains no closed path around the origin, and the same is true for the existence of a continuous branch ${\rm Log}$ of $z\mapsto\log z$.
Now given such an $\Omega$ you needed the exponential function to create $R$ from ${\rm Log}$. If you want us to "create" ${\rm Log}$ from $R$ you would have to indicate somehow what the allowable means are. Note that even in the real domain there is no natural process of this sort.
