As the title states, I would like to find a Laurent series expansion for the function, $$f(z)=\frac{7z-3}{z(z-1)},$$ in the region, $0<|z|<1$.
First, we decompose the function by partial fractions. We have, $$f(z)=\frac{7z-3}{z(z-1)}=\frac{3}{z}+\frac{4}{z-1}.$$ Based on the region, we need a Laurent series for $(z-0)$ and a Taylor series for (z-1). We note that $$\frac{3}{z}$$ is already in a Laurent series, so we can leave it alone. It remains to find a Taylor series for $$\frac{4}{z-1}.$$ We rewrite this as $$-4\cdot\frac{1}{1-z},$$ such that the Taylor series is given by, $$4\sum\limits_{n=0}^{\infty} z^n.$$ So overall we have, $$\frac{3}{z}-4\sum\limits_{n=0}^{\infty} z^n$$ as the Laurent series. I would like to know if I'm at least on the right track with my attempt.