<s>[INST] Laurent Series Expansion Problems Expand in a laurent Series :
1- $f_{1} (z) =  \frac{z^{2} - 2z +5 }{(z^{2}+1) (z-2)}$
in the ring : $1 < |z| < 2 $
2- $ f_{2} (z) =  \frac{1 }{(z-3) (z+2)}$
In : 
$i. 2 < |z| < 3
\\ ii. 0 < |z+2| < 5$
I managed to solve the second one but not sure if it is correct 
For i. $2 < |z| < 3$ :
$ \frac{-1}{5} *  \frac{1}{z(1+ \frac{2}{z}) } +  \frac{1}{5} *  \frac{1}{-3(1- \frac{z}{3}) }  
 = \frac{-1}{5} \sum_{n=0}^ \infty  (-1)^{n}  (\frac{2}{z})^{n} -  \frac{1}{15}\sum_{n=0}^ \infty    (\frac{z}{3})^{n}$
For ii. $0 < |z+2| < 5$ :
$ \frac{-1}{5} *  \frac{1}{z+2} +  \frac{1}{5} *  \frac{1}{(z+2 -5) }  
 = \frac{-1}{5} *  \frac{1}{z+2} +  \frac{1}{25} *  \frac{1}{-5 (1- \frac{Z+2}{5} ) } \\ = \frac{-1}{5} *  \frac{1}{z+2}-  \frac{1}{25} \sum_{n=0}^ \infty (\frac{z+2}{5})^{n}      $ 
 [/INST] In case 1, note that $1\lt|z|\lt2$ means that $1/|z|\lt1$ and $|z|/2\lt1$, hence the well known expansion
$1/(1-u)=\sum\limits_{n\geqslant0}u^n$, valid for every $|u|\lt1$, applied twice, yields
$$
\frac1{z^2+1}=\frac1{z^2(1+1/z^2)}=\sum\limits_{n\geqslant0}\frac{(-1)^n}{z^{2n+2}},$$
and
$$
\frac1{z-2}=-\frac12\frac1{1-z/2}=-\sum\limits_{n\geqslant0}\frac{z^n}{2^{n+1}}.
$$
The expansion of $f_1(z)$ as a Laurent series follows from the decomposition
$$
f_1(z)=\frac1{z-2}-\frac2{z^2+1}.$$
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