Find the Laurent Series for $\frac{z-2}{z+1}$ around $z = -1$
I'm not sure how to do this because it is not something with a simple numerator.
If it was something like $\frac{1}{(z-2)(z+1)}$ I would be able to do it with no issues, but I'm not sure how to attack this quotient.
Write $z-2 = z+1-3$ so we can break $$\frac{z-2}{z+1} = -\frac{3}{z+1}+1 = -3(z+1)^{-1} + 1$$and you're done.