<s>[INST] Have I found the correct Laurent series expansion? I am supposed to find the Laurent series expansion for
$$f(z) = \frac{z+1}{z^2-4}$$ in the region $1<|z+1|<3$. 
My solution:
$$w=z+1 \Leftrightarrow z=w-1 \Rightarrow f(z) = \frac{w-1+1}{(w-1)^2-4} = \frac{w}{(w+1)(w-3)}$$
Partial fractions: $$\frac{w}{(w+1)(w-3)} = \frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3}$$
We have: 
\begin{eqnarray}\frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3} &=& \frac{1}{4}\frac{1}{w} \bigg(\frac{1}{1+\frac{1}{w}}\bigg) + \frac{3}{4}\frac{1}{w}\bigg(\frac{1}{1-\frac{3}{w}}\bigg)\\ &=& \frac{1}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{(-1)}{w}\bigg]^n + \frac{3}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{3}{w}\bigg]^n\\ &=& \frac{1}{4} \sum_{n=0}^\infty \frac{(-1)^n}{w^{n+1}} + \frac{3}{4} \sum_{n=0}^\infty \frac{3^n}{w^{n+1}}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{w^{n}} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{w^n}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(z+1)^n} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{(z+1)^{n}}\end{eqnarray}
for the region $1<|z+1|<3$. It feels like there is something wrong here. Any help is greatly appreciated.
 [/INST] Seems like the second sum is good on $\mid z+1\mid\gt3$.  You should factor out $3$ instead of $w$ (in the denominator).
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