Residues of $f(z)=z^2−2z/(z+1)^2(z^2+4)$ , questions for poles of orders 2 and more Hello the exercise I had been given is finding the residues of the function $f(z)= \frac{z^2−2z}{(z+1)^2(z^2+4)}$. Here is my resolution but I'm not exactly sure about the residues at each poles. So first of all I found that $f $ has a pole of order 2 at $z = -1$ and two simples poles at$ z = +2i $ and $z = -2i$. Then I decomposed my function into partial fractions and here is what i have $f(z) = \frac{-14}{25(z+1)} + \frac{3}{5(z+1)^2} + \frac{7+i}{25(z-2i)} + \frac{7-i}{25(z+2i)}$. Since the residue is the coefficient of $z^{-1}$ in the Laurent Series of a function around a point $z_0$ (which I do have here but I don't see which $z_0$). Are the residues for $z=-1;2i;-2i$ respectively $\frac{-14}{25}$ $\frac{7+i}{25}$ and $\frac{7-i}{25}$ ? And if it is the case then I don't understand how to expand a function around a particular neighborhood $z_0$. So my first question is are my residues corrects ? And the second one is how to expand my function into Laurent Series in the neighborhood of $-1$ $2i$ and$-2i$ ? And finally in general when I have to find the residue of a pole of order m is this formula  $$Res(f,z_0) = \lim_{z\to z_0} \frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)] $$
A better method than simply finding the coefficient $a_{-1}$ of a Laurent Series ? Because I won't always be able to expand my function into a Laurent series.
Please excuse my poor English I'm a french speaker but I haven't found any help in french forums and thanks in advance for your help.
 A: Yes, your residues are correct. But you don't have to find the Laurent expansions near those three poles in order to find them.
Let me begin with the simpler ones: $\pm2i$. In the case of $2i$, you have$$f(z)=\frac{g(z)}{z-2i}\quad\text{where}\quad g(z)=\frac{z^2-2z}{(z+1)^2(z+2i)},$$and therefore\begin{align}\operatorname{res}_{z=2i}f(z)&=\operatorname{res}_{z=2i}\frac{g(z)}{z-2i}\\&=g(2i)\\&=\frac7{25}+\frac i{25}.\end{align}What I used here was the formula $\operatorname{res}_{z=a}\frac{\varphi(z)}{z-a}=\varphi(a)$, which follows from the fact that, if$$\varphi(z)=a_0+a_1(z-a)+a_2(z-a)^2+\cdots,$$then$$\frac{\varphi(z)}{z-a}=\frac{a_0}{z-a}+a_1+a_2(z-a)+\cdots.$$Now, concerning $-1$, note that$$f(z)=\frac{h(z)}{(z+1)^2}\quad\text{where}\quad h(z)=\frac{z^2-2z}{z^2+4}.$$But, near $-1$, you have$$h(z)=\frac35-\frac{14}{25}(z+1)+b_2(z+2)^2+\cdots,$$and therefore\begin{align}\operatorname{res}_{z=2i}f(z)&=\operatorname{res}_{z=2i}\frac{h(z)}{(z+1)^2}\\&=\frac{3/5}{(z+1)^2}-\frac{14/25}{z+1}+b_2+\cdots\\&=-\frac{14}{25}.\end{align}
