 Solution

If $f=z^{-n}g$ with $g(0)\ne 0$, continuity gives a small disk on which $|g(0)|/2\le |g(z)|\le 3|g(0)|/2$. Multiplying by $|z{|}^{-n}$ gives the two required bounds.

Conversely, set $g(z)=z^{n}f(z)$ on the punctured disk. The upper bound makes $g$ bounded, so the allowed removable-singularity theorem gives a holomorphic extension. The lower bound gives $|g(0)|\ge C_1>0$ by continuity. Therefore $f=z^{-n}g$ has a pole of exact order $n$. The lower bound is essential: the upper bound alone only proves order at most $n$.