find the Laurent series expansion of $f(z) = \frac{1}{(z + 1)(z +3)}$ for the region $0 \lt |{z + 1}| \lt 2$ I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$
Using partial fration f(z) can written as:
$$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3}$$
since we have from Maclurin's series,
$$\sum_0^\infty \frac{1}{1+z} = (-z)^n for |z| <1$$
so ,
$$f(z) = \frac{1}{2} [\sum (-1)^n z^n - \frac{1}{1 + (z + 2)}]$$
How to proceed from here?

we have,
$$0 < | z + 1| < 2 \implies |z| \lt 1 , \frac{|z + 1|}{2} \lt 1$$
and Using partial fraction we have,
$$f(z) = \frac{1}{2}\frac{1}{z+1}  - \frac{1}{2}\frac{1}{z+3}$$
$$\implies \frac{1}{2} \sum(-1)^n z ^n - \frac{1}{4}\frac{1}{1 + \frac{z + 1}{2}}$$
$$\implies \frac{1}{2} \sum(-1)^n z ^n - \frac{1}{4} \sum(-1)^n \frac{(z + 1)^n}{2^n}$$
