Question about logarithmic derivative In my complex analysis class, my professor mentioned something about how if $f$ and $g$ are functions defined on some domain $D\subset\mathbb{C}$, if they have identical logarithmic derivatives, then they are off by a multiplication by a constant. My question is, how can we show this given the weird behavior of $\log$ on the complex plane?
Namely, before even getting to the proof, how can we be sure that we can define $\log f$ in a way such that we can take its derivative? The function $\log$ must have a branch cut somewhere and there is no guarantee (at least for what I see) of being able to always pick a branch cut such that the values of $f$ dodge the line of discontinuity entirely. If there is a point on such a line, then I would assume that $(\log f)'$ is not even defined there. Is there a way around this that I am missing?
 A: The logaritmic derivative of a holomorphic function $f$ is the meromorphic function $\frac{f'}{f}$, with singularities at the zeros of $f$. Now, to prove the result.
One way to prove the result is as follows:
if $f\equiv 0$, the result is trivial.
So, suppose $f\not\equiv 0$
Now, let $z_0\in D$ be such that $f(z_0)\neq 0$ and there exists a disc $B(z_0,\varepsilon)\subset D$ on which $f\neq 0$.
As they have the same logarithmic derivative, $g\not= 0$ on $B(z_0,\varepsilon)$.
As $B$ is simply connected, we can define both $\ln(f)$ and $\ln(g)$ on $B$
(explicitly as $\int_{z_0}^z\frac{1}{f}$ over some path contained in  $B$: this is independent from the path by Cauchy's theorem).
Having the same logarithmic derivative, we have $\ln(f)=\ln(g)+c$ on $B$. This implies $f=e^c g$ on $B$. The identity principle does the rest, as $\frac{f}{g}=c$ on $B$ implies $\frac{f}{g}=c$ on all the domain (one can also prove the result withouth the identity principle, by considering the constant $c$ as a function of $z_0$ and proving it is a continuous locally constant function, thus constant on all of $D$)
