<s>[INST] residue $\frac{z}{(z-1)(z+1)^2}$ at $z = -1$ My attempt: $$f(z) = \frac{z}{(z-1)(z+1)^2} = \frac{1}{(z+1)^2}\cdot -z\frac{1}{1-z}$$ So at $D(-1;1)$ we have $$-z\frac{1}{1-z} = \sum_{k=0}^\infty-(z+1)^{k+1}$$ and this gives $$f(z) = \sum_{k=-1}^\infty-(z+1)^{k}.$$ So the residue at $z = -1$ is negative one but I feel like this is the wrong way of approaching this problem, because of the open disk I use. Also I think I proofed that $c_{-2} \neq 0$ because of a pole of order two here, so I'm not sure if this can be correct.
 [/INST] For each $z\in\Bbb C$ such that $|z+1|<2$, you have\begin{align}\frac z{z-1}&=1+\frac1{z-1}\\&=1-\frac1{2-(z+1)}\\&=1-\frac12\cdot\frac1{1-\frac{z+1}2}\\&=1-\sum_{n=0}^\infty\frac{(z+1)^n}{2^{n+1}}\\&=\frac12-\sum_{n=1}^\infty\frac{(z+1)^n}{2^{n+1}}\end{align}and therefore, if $z\ne-1$,\begin{align}\frac z{(z-1)(z+1)^2}&=\frac1{2(z+1)^2}-\sum_{n=1}^\infty\frac{(z+1)^{n-2}}{2^{n+1}}\\&=\frac1{2(z+1)^2}-\sum_{n=-1}^\infty\frac{(z+1)^n}{2^{n+3}}.\end{align}In particular,$$\operatorname{res}_{z=-1}\frac z{(z-1)(z+1)^2}=-\frac14.$$
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