Question: Suppose $f$ has a pole order $n$, $n\in \mathbb{Z}_{>0}, $at $z=0$ then there exists a positive constant $C$ such that $C|z|^{-n}\leq |f(z)|$ on sufficiently small punctured disk about $0$.
This is my solution so far:
Since $f$ has a pole order $n$, $f$ has Laurent expansion in the form on $\sum_{m\geq-n}c_mz^m$ for all $z$ in some punctured disk about $0$, say on $B(0,\epsilon)- \{0\}.$
Define $g(z):=z^nf(z)$, then naturally this can be extended into a (continuous) function on $B(0,\epsilon)$ by defining $g(0)=c_{-n}\not=0.$
Now since the absolue function is continuous from $\mathbb{C}\to \mathbb{R}$ and so $|g(z)|$ is continuous on $B(0,\epsilon).$ In particular, $|g(0)|$=$|c_{-n}|>0.$ Then by choosing $\epsilon=\frac{1}{2}|c_{-n}|,$ we have $\exists\delta>0$ such that $|g(z)|\geq \frac{1}{2}|c_{-n}|$ on $B(0,\delta).$
Then on $B(0,\delta)-\{0\}$, we have the required inequality by setting $C=\frac{1}{2}|c_{-n}|.$
I was wondering if there are any flaws with my solution, the hint tells me to consider removable singularity but I did not use it much and so I was wondering if there is a slicker way of doing it. Many thanks!
Your solution is fine. However, there's no need to consider the Laurent expansion--you can just directly look at $g(z)=z^nf(z)$ and observe that its singularity at $z=0$ is removable with $g(0)\neq 0$ since $f$'s pole there has order $n$. The rest of the argument is then the same as yours.