Consider the function $$f(z) = {1 \over (z-i)(z+i)}$$
with a Laurent series expansion at $z_0=i$ on a domain $\;\Omega=\left\{z\in \mathbb{C}:2\lt\left|z-i\right|\right\}$ $$\begin{eqnarray}f(z)={1 \over (z-i)(z+i)}={1 \over (z-i)}{1 \over (z+i)}={1 \over (z-i)}{1\over(z-i)+2i} \\={1 \over (z-i)}{1 \over (z-i)}{1\over1-\left(-{2i\over z-i}\right)}\\={1 \over (z-i)^2}\sum_{n=0}^\infty\left(-{2i\over z-i}\right)^n\\=\sum_{n=0}^\infty (-2i)^n\left({1\over z-i}\right)^{n+2}\\=\sum_{n=-\infty}^{-2}\left(\frac i2\right)^{n+2}(z-i)^n\end{eqnarray}$$
Just to make sure I'm getting everything correct. Obviously we have a singularity at $z=i$, however cannot determine its type, as $i\notin \Omega$. Is it true that we can't determine the type of this singularity on the restricted domain as it doesn't lie in this domain?
Set $z=i+w$, then
$$f(z) = g(w)={1 \over w(w+2i)}=\frac{1}{2iw (1+(w/2i))}=\frac{1}{2iw}\left(1-(w/2i)+(w/2i)^2-(w/2i)^3+\cdots\right)$$
So $w=0$($z=i$) is a pole of $g(w) (f(z))$.