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?

It should rather be
$$f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \frac{1}{1+\frac{(z+1)}{2}}$$
$$\implies f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \left(\frac{z+1}{2}\right)^k.$$
