= Solution
Put $\zeta=z-3$ and apply the <geometric series>:
$$
\frac1{z-2i}=\frac1{3-2i}\frac1{1+\zeta/(3-2i)}=\sum_{n=0}^\infty\frac{(-1)^n}{(3-2i)^{n+1}}(z-3)^n.
$$
Thus
$$
\boxed{a_n=\frac{(-1)^n}{(3-2i)^{n+1}},\qquad |z-3|<\sqrt{13},\qquad R=\sqrt{13}.}
$$
The radius is the distance from the expansion centre $3$ to the nearest singularity $2i$, not the distance of that <pole> from the origin. On the boundary circle the series terms all have magnitude $1/\sqrt{13}$, so they do not approach zero: this particular series diverges at every boundary point.
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