What's the Laurent series expansion of $\frac{z-2}{z+1}$ at $z=-1$?
2026-03-31 21:51:22.1774993882
Laurent Series of $(z-2)/(z+1)$ at $z=-1$
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We have
$$\frac{z-2}{z+1} = \frac{(z + 1) - 3}{z+1} = 1-\frac{3}{z+1} = \frac{1}{(z+1)^0} + \frac{-3}{(z+1)^1}.$$
So $$\frac{z-2}{z+1} = \sum_{n=-\infty}^\infty a_n (z+1)^n$$ where $a_0 = 1$, $a_{-1} = -3$ and $a_j = 0$ otherwise.