= Solution
Put $w=z-1$ and $a=1-i$. <Partial fraction decomposition> gives
$$
\frac1{(z-i)(z+3)}=\frac1{3+i}\left(\frac1{w+a}-\frac1{w+4}\right).
$$
Both terms admit a <geometric series> at $w=0$. The local <Laurent series> is actually a <Taylor series>, since the center is regular:
$$
\boxed{\frac1{(z-i)(z+3)}=\frac1{3+i}\sum_{m=0}^{\infty}(-1)^m\left[\frac1{(1-i)^{m+1}}-\frac1{4^{m+1}}\right](z-1)^m,\quad |z-1|<\sqrt2.}
$$
The nearest <pole> is $z=i$, a distance $|1-i|=\sqrt2$ from the center; the other <pole> is at distance $4$. Thus the requested local <radius of convergence> is $\boxed{R=\sqrt2}$.
For completeness, specifying a <Laurent series> without a region can also mean either of the other maximal annuli. Expanding the first fraction in inverse powers and the second in positive powers gives
$$
\frac1{3+i}\left[\sum_{m=0}^{\infty}\frac{(-a)^m}{w^{m+1}}-\sum_{m=0}^{\infty}\frac{(-1)^mw^m}{4^{m+1}}\right],\qquad \sqrt2<|w|<4.
$$
Expanding both fractions in inverse powers instead gives
$$
\frac1{3+i}\sum_{m=0}^{\infty}\frac{(-a)^m-(-4)^m}{w^{m+1}},\qquad |w|>4.
$$
These are different <Laurent series> on different annuli, rather than extensions of the local <Taylor series> beyond its <radius of convergence>.
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