= Solution
If $f=z^{-n}g$ with $g(0)\ne0$, <continuity> gives a small disk on which $|g(0)|/2\le|g(z)|\le3|g(0)|/2$. Multiplying by $|z|^{-n}$ gives the two required bounds.
Conversely, set $g(z)=z^nf(z)$ 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 $n$.
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