<s>[INST] laurent series of $\frac{z^2-2z+2}{(z-1)(z^2-2z-3)}$ around $z-1$ for $0<|z-1|<2$ I want to find the Laurent series of $L=\frac{z^2-2z+2}{(z-1)(z^2-2z+3}$ around $(z-1)$
Here, I used the fact that 
$L=\frac{5}{8(z-3)}+\frac{5}{8(z+1)}-\frac{1}{4(z-1)}$
When $|z-1|<2$, we have
$L=\frac{-1}{4(z-1)}+\frac{5}{8} \frac{1}{(z-1)-2} +\frac{5}{8( (z-1)+2)}$
$=\frac{-1}{4(z-1)} +\frac{5}{-16}\frac{1}{1- \frac{(z-1)}{2}} + \frac{5}{16} \frac{1}{1+ \frac{(z-1)}{2}} =-\frac{1}{4(z-1)}+ \frac{5}{16} \sum_{k=0}^\infty \frac{(z-1)^k}{2^k}(-1+(-1)^k)=-\frac{1}{4(z-1)}+\frac{5}{16} \sum_{n=0}^\infty \frac{(z-1)^{2n+1}}{2^{2n+1}} (-2)$
However,in my book, Elements d'analyse complexe by Real Gelinas,
the answer for 0<|z-1|<2 is 
$\frac{-1}{4} (z-1 + \frac{1}{z-1} )\sum_{n=0}^\infty (\frac{z-1}{2n})^{2n}.$
Hence, I am missing a term.
Where is my mistake?
 [/INST] Note : Since you've made a mistake in your expressions while solving (the one mentioned in the comments) I'll go over a more straight-forward and common approach, by demonstrating its all steps.
The trick is to form the expression that is asked to form the Laurent series around, which in that specific case is : $z-1$. This is what I'm going to demonstrate down below : 
$$L(z) = \frac{z^2-2z+2}{(z-1)(z^2-2z-3)}= \frac{(z-1)^2+1}{(z-1)(z+1)(z-3)}$$
$$=$$
$$\frac{(z-1)^2+1}{(z-1)(z-1+2)(z-1-2)} = \frac{(z-1)^2+2}{(z-1)(z-1)(z-1)(1+\frac{2}{z-1})(1-\frac{2}{z-1})}$$
$$=$$
$$\frac{(z-1)^2+1}{(z-1)^3(1+\frac{2}{z-1})(1-\frac{2}{z-1})}$$
Now, we know the following simple geometric series : 
$$\frac{1}{1-w}\sum_{n=0}^\infty w^n \quad |w|<1$$
$$\frac{1}{1+w} = \sum_{n=0}^\infty (-1)^nw^n \quad |w|<1$$
Applying these for $\frac{2}{z-1} = w$ :
$$[(z-1)^2+1] \frac{1}{(z-1)^3}\sum_{n=0}^\infty(-1)^n2^n(z-1)^{-n}\sum_{n=0}^\infty 2^n(z-1)^{-n} $$
$$=$$
$$\bigg[\frac{1}{z-1}+\frac{1}{(z-1)^3}\bigg]\sum_{n=0}^\infty(-1)^n2^n(z-1)^{-n}\sum_{n=0}^\infty 2^n(z-1)^{-n} $$
Can you now combine all of these into one expression and yield your results ? For the relation of $|z-1|$ given, it's easy to spot since you'd want $|\frac{2}{z-1}| < 1$ and on the same time $z-1\neq 0$ which means given your first condition that $|z-1| > 0$. Combining these you get : $0<|z-1|<2$.
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