Laurent series for $\frac{z - 2}{z + 1}$
by Hidenori
Proposition
Find a Laurent series for $\frac{z - 2}{z + 1}$ centered at $z = -1$ and the region in which it converges.
Solution
\[\begin{align*} \frac{z - 2}{z + 1} &= \frac{(z + 1) - 3}{z + 1} \\ &= 1 - \frac{3}{z + 1} \\ &= \frac{-3}{z + 1} + 1. \end{align*}\]This is a Laurent series with $c_{-1} = -3, c_0 = 1$. This converges if $z \ne -1$.
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