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.
 A: 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,$$\operatorname{res}_{z=-1}\frac z{(z-1)(z+1)^2}=-\frac14.$$
A: For a function with low and obvious pole degrees you can use the limit definition.
Recall the residue of an $n$ degree pole at $c$ of $f$ is $$\frac{1}{(n-1)!}\lim_{z\to c}\frac{d^{n-1}}{dz^{n-1}}[(z-c)^nf(z)]$$
For your function $$\frac{z}{(z-1)(z+1)^2}$$ this has a simple pole at $1$ and a pole of order 2 at $-1$. Hence by the formula at $-1$ the residue is nothing but
$$\frac1{(2-1)!}\lim_{z\to -1}\frac{d}{dz}\left[{\color{red}{(z+1)^2}}\cdot \frac{z}{(z-1){\color{red}{(z+1)^2}}}\right]=\lim_{z\to-1} {-1\over(-1 + z)^2}=-\frac14$$
Edit: Ok i know that this has already been answered but w/e i'll still put my sereis answer here. We have $$\frac{z}{(z-1)(z+1)^2}={\color{blue}{(z+1)^{-2}}}\cdot \frac z{z-1}$$
Blue is already its own laurent series at $-1$, so we just need to find the series of the latter. Taylor expanding at -1 gives
$$(z+1)^{-2}\cdot \left( \frac12-\frac14 (z+1) - \frac18 (z+1)^2 - \frac1{16}(z+1)^3+... \right)$$
Notice that if we multiply it out, $(z+1)^{-2}\cdot -\frac14 (z+1)$ gives the negative one power term, implying the residue at that -1 is $-\frac14$.
A: Set $z=-1+h.$ For $0<|h|<1,$
$$\begin{align}\frac z{(z-1)(z+1)^2}&=\frac{-1+h}{(-2+h)h^2}\\&=\frac{1-h}{2h^2(1-\frac h2)}\\&=\frac{1-h}{2h^2}\left(1+\frac h2+o(h)\right)\\&=\frac1{2h^2}\left(1-\frac h2+o(h)\right).
\end{align}$$
The residue of $\frac z{(z-1)(z+1)^2}$ at $z=-1$ is the coefficient of $\frac1h$ in this developement, i.e. $-\frac14.$
