<s>[INST] Laurent series of $f(z)=\frac{4z-z^2}{(z^2-4)(z+1)}$ in different annulus Given $f(z)=\dfrac{4z-z^2}{(z^2-4)(z+1)}$ 
I need to find the Laurent series in the annulus: $A_{1,2}(0),\;A_{2,\infty}(0),\;A_{0,1}(-1)$
I found the following partial fractions:
$f(z)=\dfrac{-3}{(z+2)}+\dfrac{1}{3(z-2)}+\dfrac{5}{3(z+3)}$,
the power series of these fractions are:
$\dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $
$\dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $
$\dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( -z \right)^n} $
and the principle parts are:
$\dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $
$\dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $
$\dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( \frac{1}{-z} \right)^n} $
In the first annuli I take the principle part only of $\dfrac{5}{3(z+1)}$, in the second annuli I take the principle part of all fraction. About the third one, I have $0<\vert z-1\vert<1$, I denoted $w=z-1$ and then I took the power series for all fractions and simply switched the $w$ back to $z-1$ at the end. Is it the right way of doing it?
I received $\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n - \frac{1}{6}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n + \frac{5}{3}\sum_{n=0}^\infty \left( 1-z \right)^n}$
 [/INST] There's a typo in your partial fraction decomposition; it should be$$\frac{4z-z^2}{(z^2-4)(z+1)}=-\frac3{z+2}+\frac1{3(z-2)}+\frac5{3(z+1)}.$$If $z\in A_{1,2}(0)$, then:


*

*$\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=-\frac32\sum_{n=0}^\infty\frac{(-1)^n}{2^n}z^n$;

*$\displaystyle\frac1{3(z-2)}=-\frac16\frac1{1-\frac z2}=-\frac16\sum_{n=0}^\infty\frac{z^n}{2^n}$;

*$\displaystyle\frac5{3(z+1)}=\frac53\frac1{1+z}=-\frac53\sum_{n=-\infty}^{-1}(-1)^nz^n$.


In $A_{2,\infty}(0)$, the third expansion remains the same, but now we have:


*

*$\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=\frac32\sum_{n=-\infty}^{-1}\frac{(-1)^n}{2^n}z^n$;

*$\displaystyle\frac1{3(z-2)}=-\frac16\frac1{1-\frac z2}=\frac16\sum_{n=-\infty}^{-1}\frac{z^n}{2^n}$.


In the case of $A_{0,1}(-1)$, you write\begin{align}\frac{4z-z^2}{z^2-4}&=-1+\frac1{z-2}+\frac3{z+2}\\&=1+\frac1{(z+1)-3}+\frac3{(z+1)+1}.\end{align}From this, you get the Taylor series of $\dfrac{4z-z^2}{z^2-4}$ in $D(1,1)$ and then, when you divide everything by $z+1$, you get the Laurent series that you're after.
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