= Solution
If $z_0\ne0$, factor out $z_0$ and use the <geometric series>:
$$
\frac1{z_0+z}=\frac1{z_0}\frac1{1+z/z_0}=\sum_{n=0}^\infty\frac{(-1)^n}{z_0^{n+1}}z^n,\qquad |z|<|z_0|.
$$
Thus
$$
\boxed{a_n=\frac{(-1)^n}{z_0^{n+1}},\qquad R=|z_0|.}
$$
The <radius of convergence> is the distance to the nearest singularity, the <simple pole> at $z=-z_0$. The case $z_0=0$ must be treated separately: the function is $1/z$, has no <Taylor series> about zero and is already its own <Laurent series> on $0<|z|<\infty$.
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