Solution
= Solution
Put $w=z-4i$. Factor out the nonzero constant in the denominator and use the <geometric series>:
$$
\frac1{3-z}=\frac1{3-4i-w}
=\frac1{3-4i}\frac1{1-w/(3-4i)}
=\boxed{\sum_{n=0}^\infty\frac{(z-4i)^n}{(3-4i)^{n+1}}}.
$$
This <Taylor series> converges when $|z-4i|<|3-4i|$. The <simple pole> at $z=3$ is the only finite obstruction to analyticity, whose distance from the expansion centre is five. Hence the <radius of convergence> is
$$
\boxed{R=5}.
$$