[proofplan]
We analyse the logarithmic derivative $f'/f$ locally near each zero and pole of $f$ by factoring out the leading-order term of the Laurent expansion. Near a zero of order $m$, write $f(z) = (z - z_0)^m g(z)$ with $g$ holomorphic and non-vanishing, then compute $f'/f$ directly. Near a pole of order $m$, write $f(z) = (z - z_0)^{-m} h(z)$ with $h$ holomorphic and non-vanishing, and repeat the computation. Away from zeros and poles, $f'/f$ is holomorphic, so the only singularities of $f'/f$ are simple poles at the zeros and poles of $f$.
[/proofplan]
[step:Show that $f'/f$ is meromorphic on $\Omega$]
Since $f \in \mathcal{M}(\Omega)$ and $f \not\equiv 0$, the zeros and poles of $f$ form a discrete subset of $\Omega$ (by the [Identity Theorem](/theorems/???) for meromorphic functions). On $\Omega \setminus \{$zeros and poles of $f\}$, the function $f$ is holomorphic and non-vanishing, so $f'/f$ is holomorphic there. It remains to determine the behaviour of $f'/f$ at the zeros and poles of $f$, which we do in the subsequent steps.
[/step]
[step:Compute $f'/f$ near a zero of order $m$]
Let $z_0 \in \Omega$ be a zero of $f$ of order $m \geq 1$. Then there exists a neighbourhood $U \subset \Omega$ of $z_0$ and a holomorphic function $g: U \to \mathbb{C}$ with $g(z_0) \neq 0$ such that
\begin{align*}
f(z) = (z - z_0)^m g(z) \quad \text{for all } z \in U.
\end{align*}
Differentiating using the product rule:
\begin{align*}
f'(z) = m(z - z_0)^{m-1} g(z) + (z - z_0)^m g'(z).
\end{align*}
For $z \in U \setminus \{z_0\}$, dividing by $f(z) = (z - z_0)^m g(z)$:
\begin{align*}
\frac{f'(z)}{f(z)} = \frac{m}{z - z_0} + \frac{g'(z)}{g(z)}.
\end{align*}
Since $g(z_0) \neq 0$ and $g$ is holomorphic on $U$, the function $g'/g$ is holomorphic on a neighbourhood of $z_0$. Therefore $f'/f$ has a simple pole at $z_0$ with residue $+m$.
[guided]
We want to understand the local behaviour of $f'/f$ near a zero $z_0$ of order $m$. The key idea is to isolate the vanishing factor: since $z_0$ is a zero of order $m$, we can write $f(z) = (z - z_0)^m g(z)$ where $g: U \to \mathbb{C}$ is holomorphic on some neighbourhood $U$ of $z_0$ and $g(z_0) \neq 0$. This factorisation exists by the definition of the order of a zero — the Taylor series of $f$ at $z_0$ begins at the $m$-th term.
Now we differentiate using the product rule:
\begin{align*}
f'(z) = m(z - z_0)^{m-1} g(z) + (z - z_0)^m g'(z).
\end{align*}
For $z \neq z_0$, we divide by $f(z) = (z - z_0)^m g(z)$:
\begin{align*}
\frac{f'(z)}{f(z)} = \frac{m(z - z_0)^{m-1} g(z) + (z - z_0)^m g'(z)}{(z - z_0)^m g(z)} = \frac{m}{z - z_0} + \frac{g'(z)}{g(z)}.
\end{align*}
Why is $g'/g$ holomorphic near $z_0$? Because $g$ is holomorphic and $g(z_0) \neq 0$, so $g$ does not vanish in some neighbourhood of $z_0$ (by continuity), and the quotient of two holomorphic functions with non-vanishing denominator is holomorphic.
The decomposition $f'/f = m/(z - z_0) + g'/g$ is exactly the Laurent expansion of $f'/f$ near $z_0$: the principal part is $m/(z - z_0)$, which is a simple pole, and the residue — the coefficient of $(z - z_0)^{-1}$ — is $+m$.
[/guided]
[/step]
[step:Compute $f'/f$ near a pole of order $m$]
Let $z_0 \in \Omega$ be a pole of $f$ of order $m \geq 1$. Then there exists a neighbourhood $V \subset \Omega$ of $z_0$ and a holomorphic function $h: V \to \mathbb{C}$ with $h(z_0) \neq 0$ such that
\begin{align*}
f(z) = (z - z_0)^{-m} h(z) \quad \text{for all } z \in V \setminus \{z_0\}.
\end{align*}
Differentiating using the product rule:
\begin{align*}
f'(z) = -m(z - z_0)^{-m-1} h(z) + (z - z_0)^{-m} h'(z).
\end{align*}
For $z \in V \setminus \{z_0\}$, dividing by $f(z) = (z - z_0)^{-m} h(z)$:
\begin{align*}
\frac{f'(z)}{f(z)} = \frac{-m}{z - z_0} + \frac{h'(z)}{h(z)}.
\end{align*}
Since $h(z_0) \neq 0$ and $h$ is holomorphic on $V$, the function $h'/h$ is holomorphic on a neighbourhood of $z_0$. Therefore $f'/f$ has a simple pole at $z_0$ with residue $-m$.
[guided]
Now consider a pole $z_0$ of $f$ of order $m$. By definition, this means $f$ has a Laurent expansion near $z_0$ whose most singular term is $(z - z_0)^{-m}$. Equivalently, the function $(z - z_0)^m f(z)$ extends to a holomorphic function at $z_0$ with a non-zero value there. So we write $f(z) = (z - z_0)^{-m} h(z)$ where $h: V \to \mathbb{C}$ is holomorphic on a neighbourhood $V$ of $z_0$ with $h(z_0) \neq 0$.
The computation mirrors the zero case. Differentiating:
\begin{align*}
f'(z) = -m(z - z_0)^{-m-1} h(z) + (z - z_0)^{-m} h'(z).
\end{align*}
Dividing by $f(z) = (z - z_0)^{-m} h(z)$ for $z \neq z_0$:
\begin{align*}
\frac{f'(z)}{f(z)} = \frac{-m(z - z_0)^{-m-1} h(z) + (z - z_0)^{-m} h'(z)}{(z - z_0)^{-m} h(z)} = \frac{-m}{z - z_0} + \frac{h'(z)}{h(z)}.
\end{align*}
Again, $h'/h$ is holomorphic near $z_0$ since $h(z_0) \neq 0$. The principal part of $f'/f$ at $z_0$ is $-m/(z - z_0)$, giving a simple pole with residue $-m$.
Notice the sign difference: zeros contribute $+m$ and poles contribute $-m$. This is intuitive — the logarithmic derivative $(\log f)' = f'/f$ measures the "rate of growth" of $f$, and near a zero $f$ is growing (from zero), while near a pole $f$ is blowing up in the sense that $1/f$ has a zero.
[/guided]
[/step]
[step:Conclude that $f'/f \in \mathcal{M}(\Omega)$]
The singularities of $f'/f$ in $\Omega$ occur precisely at the zeros and poles of $f$, which form a discrete subset of $\Omega$. At each such point, we have shown that $f'/f$ has a simple pole (with residue $\pm m$). Away from these points, $f'/f$ is holomorphic. A function that is holomorphic except for isolated poles is meromorphic, so $f'/f \in \mathcal{M}(\Omega)$.
[/step]
