 Solution

A function has a pole of order $n$ at $a$ when, on a punctured neighbourhood,Equivalently, $(z-a{)}^{n}f(z)$ has a removable singularity whose extension is nonzero at $a$. After shrinking the neighbourhood so $h\ne 0$, its reciprocal $1/f$ extends with a zero of order $n$. In particular, $|f(z)|\to \infty$ at a pole.