Solution
= Solution
A \b[<pole>] of order $n\geq1$ means
$$
\boxed{f(z)=(z-z_0)^{-n}h(z),\qquad h\text{ holomorphic near }z_0,\quad h(z_0)\ne0.}
$$
Equivalently, the <Laurent series> has a finite nonzero <principal part>, with lowest power exactly $(z-z_0)^{-n}$. Thus $(z-z_0)^nf(z)$ extends holomorphically with nonzero value at $z_0$; using a smaller positive power would not remove the singularity.