Solution
= Solution
For $|z|<1$, differentiate the <geometric series> to obtain $(1-t)^{-2}=\sum_{k\ge0}(k+1)t^k$. Since $(z-i)^2=-(1-z/i)^2$,
$$
\boxed{\frac1{z(z-i)^2}=-\sum_{k=0}^{\infty}(k+1)(-i)^kz^{k-1}=-\frac1z+2i+3z-4iz^2-5z^3+\cdots.}
$$
The convergence region is precisely $0<|z|<1$, the specified punctured disk; the other <pole> is on its boundary at $i$.