If $f$ is meromorphic in some region, why can't $\frac{f'}{f}$ have any essential singularities in that region? It is stated in the Wikipedia article on the argument principle that the only singularities of $\frac{f'}{f}$ are the zeros and poles of $f$ itself.
It is not at all clear that there are no essential singularities; suppose $f$ has an essential singularity - surely so would its derivative, and so how can it be that the quotient merely has a tame pole, instead of an essential singularity.
My only thought is this:
$$\frac{f'(z)}{f(z)}=\frac{d}{dz}\ln(f(z))$$
And so the only essential singularities of the quotient would be the essential singularities of $\ln$, which is essentially singular (is this correct terminology?) at $z=0$, but at $z=0$ the quotient has a pole, therefore we only have a pole. This however is massively unrigorous! Is this correct? How does one prove it formally?
 A: $f$ is meromorphic in a domain $D \subset \Bbb C$ if there is a (possibly empty) set $P \subset D$ such that

*

*every point in $P$ is isolated in $D$,

*$f$ is holomorphic in $D \setminus P$, and

*$f$ has a pole at every point in $P$.

Now assume that $f$ is not identically zero, and let $Z$ be the (possibly empty) set of the zeros of $f$. (We know from the identity principle that every point in $Z$ is isolated in $D$.)
The $h = f'/f$ is holomorphic in $D \setminus (P \cup Z)$.
At a $k$-fold pole of $f$ at $a \in P$ is
$$
 f(z) = (z-a)^{-k} g(z)
$$
where $g$ is holomorphic and non-zero in a neighborhood of $z=a$. Therefore
$$
 h(z) = \frac{f'(z)}{f(z)} = \frac{-k}{z-a} + \frac{g'(z)}{g(z)}
$$
in a neighborhood of $z=a$, so that $h$ has a simple pole at $z=a$.
Similarly, at a $l$-fold zero of $f$ at $b \in Z$ is
$$
 f(z) = (z-b)^{l} h(z)
$$
where $h$ is holomorphic and non-zero in a neighborhood of $z=b$. Therefore
$$
 h(z) = \frac{f'(z)}{f(z)} = \frac{l}{z-b} + \frac{h'(z)}{h(z)}
$$
in a neighborhood of $z=b$, so that $h$ again has a simple pole at $z=b$.
So the only singularities of $f'/f$ occur at the zeros and poles of $f$, and these singularities are poles.
