As title suggests, I want to find a Laurent series expansion for $$\frac{1}{z^2-4z+3}=\frac{1}{(z-3)(z-1)}=\frac{1}{2}\frac{1}{z-3}-\frac{1}{2}\frac{1}{z-1}$$ about $z=0$ with $|z|<1$. We have singularities at $z=1$, and $z=3$. I am confused about whether I want a Taylor series for both terms. That is, is the series expansion given by the following:
$$\frac{1}{2}\sum_{0}^{\infty}\frac{z^n}{3^n}-\frac{1}{2}\sum_{0}^{\infty}z^n?$$ I'm somewhat confused about the region. Both singularities lie outside the center, so they should both be Taylor series expansions, but I'm not so sure.