Solution
= Solution
A function has a <pole> of order $n$ at $a$ when, on a punctured neighbourhood,
$$
\boxed{f(z)=\frac{h(z)}{(z-a)^n},\qquad h\text{ holomorphic near }a,\quad h(a)\ne0}.
$$
Equivalently, $(z-a)^n f(z)$ has a <removable singularity> whose extension is nonzero at $a$. After shrinking the neighbourhood so $h\ne0$, its reciprocal $1/f$ extends with a zero of order $n$. In particular, $|f(z)|\to\infty$ at a <pole>.