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.
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}