Laurent series question 2

Question

Expand $f(z)=\frac{1}{(z+1)(z+3)}$ in a Laurent series valid for $0<|z+1|<2$

I've attempted to make a substitution of $z+1=w$ but I'm not sure how to show that $w>0$

Can somebody please help.

Answer 1

Hint:

Justify and fill in details in the following:

$$\frac1{(z+1)(z+3)}=\frac12\left(\frac1{z+1}-\frac1{z+3}\right)=\frac1{2(z+1)}-\frac14\frac1{1+\frac{z+1}2}=$$

$$=\frac1{2(z+1)}-\frac14\left(1-\frac{z+1}2+\frac{(z+1)^2}4-\ldots+\frac{(-1)^n(z+1)^n}{2^n}+\ldots\right)=$$

$$=\frac12(z+1)^{-1}-\sum_{n=0}^\infty\frac{(-1)^n(z+1)^n}{4\cdot2^n}$$