Here's the definition.
Let $X$ be a Riemann surface and $Y$ be an open subset of $X$.
A meromorphic function on $Y$ is a holomorphic function $f\colon Y' \rightarrow \mathbb{C}$ satisfying the following conditions, where $Y'\subset Y$ is a open subset.
(1) Every point $p\in Y – Y'$ is an isolated point.
(2) For every point $p\in Y – Y'$, $\lim_{x\rightarrow p} |f(x)| = \infty$.
The points of $Y – Y'$ are called the poles of $f$.
Then he stated in a remark:
Let $(U, z)$ be a coordinate neighborhood of a pole $p$ of $f$ with $z(p) = 0$.
Then $f$ can be expanded in a Laurent series $f = \sum_{n = -k}^{\infty} c_n z^n$ in a neighborhood of $p$.
Why is this so? In other words, why $p$ cannot be an essential singularity?
Best Answer
As pointed out in the commentaries, assumption $(2)$ ensures that the singularity is a pole.
The statement is local, so let's say $f$ is a holomorphic function on $U\setminus{0}$, where $U\subset\mathbb{C}$ is a neighborhood of the origin in the complex plane, and that $\lim_{z\to 0}{|f(z)|} = \infty$. This allows us to assume, that $f$ doesn't vanish on $U\setminus{0}$ (by possibly shrinking $U$). Thus $g = 1/f$ is holomorphic on $U\setminus{0}$ and bounded near zero. Therefore is has a unique holomorphic extension to $U$, obviously given by $g(0) = 0$ and this zero point is isolated, hence it has a finite order of vanishing. But this order of vanishing is, by definition, the order of the pole of $1/g = f$, i.e. $f$ has a pole of finite order in $0$ (and especially not an essential singularity).