Suppose we have a non-zero holomorphic function $f$ in a simply-connected region $D$. I know that there exists $g=\int_{s_0}^s\frac{f'}{f}dz$ s.t $e^g=f$. So this is the "logarithm" of $f$ in $D$.
But what is the difference been this "logarithm" and taking some branch of the complex logarithm and composing it with f i.e. $Log(f(s))$. For instance if we use the principal branch we have to ensure that Range($f$) does not include negative reals.
So is $g$ supposed to be just some branch cut composed with $f$ where the range of f lies completely in the domain of this branch cut. If so how do we know that such a branch cut exists?
In general, the answer is no (as in the case mentioned $\Im \log f$ which is an argument function for $f$ would be bounded) and $g = \log f$ takes different values at values of $f(s)$ that are equal. Note that since $\Re g=\log |f|$, the real part is the same, just the imaginary part of $g$ is different by integral multiples of $2\pi$
A simple example is $e^z$ on the full plane, which has as one possible logarithm (unsurprisingly) $z$, hence the argument of $e^z$ "rolls" in the plane
If we know that the image of $g$ is contained in a domain avoiding a ray from $0$ (for example $f(z) \ne -x, x \ge 0$ or easier sometimes $\Re f >0$) then yes we can construct a logarithm coming from a unique branch, hence having bounded imaginary part. In the case $\Re f >0$ for example $|\arg f|=|\Im \log f| < \frac{\pi}{2}$ for the "principal logarithm of $f$ which immediately for example gives nice results like $\Re \sqrt f >0$ for the principal branch of the square root etc