= Solution
Clear the apparent negative power of $z$ to obtain the <rational function>
$$
h(z)=\frac{-2iz}{z^2-2iaz-1}.
$$
Let $\Delta=\sqrt{a^2-1}>0$. Its denominator factors as $(z-z_-)(z-z_+)$, with
$$
\boxed{z_-=i(a-\Delta),\qquad z_+=i(a+\Delta).}
$$
These are distinct <simple poles>. Since $(a-\Delta)(a+\Delta)=1$ and $a+\Delta>1$, we have $0<a-\Delta<1$: only $z_-$ lies inside the <unit circle>. The apparent singularity at $z=0$ in the original expression is a <removable singularity>, with the continued value $h(0)=0$; it is not another pole.
Using the <residue> formula from part (a), with denominator derivative $2z-2ia$, gives
$$
\boxed{\operatorname{Res}_{z=z_-}h=\frac{-2iz_-}{2z_--2ia}=\frac{i(a-\Delta)}{\Delta}.}
$$
There are no other finite singularities of the continued <rational function>.
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