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?

For the first term, it's already done:  $$\dfrac 12\dfrac 1{z+1}$$ on $\mid z+1\mid\gt0$.
For the other one,  $$\dfrac 1{z+3}=\dfrac 1{(z+1)+2}=\dfrac 1{2--(z+1)}=\dfrac 12\dfrac 1{1--\dfrac {z+1}2}=\dfrac 12\sum_{n\ge0}(-\dfrac {z+1}2)^n$$ on $\mid z+1\mid\lt2$.
So to finish we have $$\dfrac 12\dfrac 1{z+1}-\dfrac 14\sum_{n\ge0}(-\dfrac{z+1}2)^n$$, on the overlap $0\lt\mid z+1\mid\lt2$.
