How is the integral of $\frac{f^\prime}{f}$ being chosen for proofs of the Complex Logarithm and Roots? I have been working with some friends to make sense of a proof in Ulrich's Complex Made Simple, which follows below:

Corollary 4.15. Any nonvanishing holomorphic function in a simply connected set has a holomorphic logarithm. That is, if V is simply connected, $f \in H(V)$ and $f$ has no zero in $V$ then there exists $L \in H(V)$ with $$ e^L = f$$ in V.


Proof. It follows from Theorem 4.0 [integral of a holomorphic function over an open set is 0 everywhere if and only if the function is the derivative of some holomorphic function in the open set] and Theorem 4.14 [Cauchy's Theorem for simply connected sets] that there exists $F \in H(V)$ such that
$$F^\prime = \frac{f^\prime}{f}. [\text{WHAT IS THIS!?!}]$$
Now the chain rule show that $(fe^{-F})^\prime =0$, so $fe^{-F}$ is constant. Setting $L = F + C$ for a suitable $c \in \mathbb{C}$ we obtain $fe^{-L} = 1$. $\square$

Now, everything about the proof and it's result makes sense to us, including how you can show, using this idea, that any nonvanishing holomorphic function in a simply connected set has a holomorphic nth root.
The only thing we do not understand is how they chose their function. I have seen this pop up in every explanation or proof of this theorem, but I do not believe it is ever addressed as to where this comes from. The only time I have seen it be addressed is in reference to the fact that $f^\prime$ and $f$ are both holomorphic over the set (which we get and that makes sense to us), but never where they come up with the idea to use $\frac{f^\prime}{f}$ as their function for $F^\prime$.
Any help clarifying this would be fantastic!
 A: It's the logarithmic derivative.
A: Honestly, I think the hardest part of the problem is brushed over the most.
Background: Recall a complex number $z$ can be written as $z=re^{i\theta}$. The angle $\theta$ is the argument and the principle value has the form $Arg\,z\in(-\pi,\pi].$ It can then be shown that
$
\log z=\ln(r)+i(Arg\,z+2k\pi),
$
which is multi-valued. For simplicity, set $k=0$. Then this function fails to be continuous (so not analytic) along the non-positive real axis because the Arg term flips from $-\pi$ to $\pi$, or vice versa.
All of this is to say the following. If the simply connected set $V$ avoids the non-positive real axis, then the logarithm (for any $k$) is analytic inside $V$.
However, suppose $V$ just avoids $z=0$, but may cross through the non-positive real axis. Choose a half-line from the origin that does not cross through $V$. Then we use this half-line as the "bad points," and define the log in such a way that the Arg creates discontinuities at points other than at $\pi$.
For example, suppose $V$ is the open disk of radius $1$, centered at $z=-2$. Then the normal log is not analytic inside this region since it crosses through the non-positive axis. However, define $\log z=\ln(r)+i(Arg^*z +2k\pi)$, where $Arg^*\,z\in (-\frac{\pi}{2},\frac{3\pi}{2}]$. This log fails to be analytic along the non-positive imaginary axis, but is fine in the region $V$.
In your problem: The hardest part of the problem is establishing that such a version of log exists for $V$. Since $f$ does not vanish inside $V$, any such definition of $\log f(z)$ that is used will satisfy $f(z)\neq 0$ so, based on what I wrote above, there exists a half-line from the origin such that $f(z)$ avoids this half-line for all $z\in V$.
With all of this background work, it is then pretty routine to calculate the derivative of $\log f(z)$ if we assume $f(z)$ is analytic and $\log$ is analytic inside the region.
EDIT: I guess the proof doesn't really need my analysis, but my analysis provides another interpretation behind why a holomorphic $\log f(z)$ exists.
