Expand the function $$f(z) = \frac{1}{(z + 1)(z + 3)}$$ in a Laurent series valid for $1 < |z| < 3$
My attempt:
$$\frac{1}{(z + 1)(z + 3)}=\frac{1}{4}.\frac{1}{1+z}-\frac{1}{4}.\frac{1}{3+z}$$
$$=\frac{1}{4}.\frac{1}{1-(-z)}-\frac{1}{4}.\frac{1}{3-(-z)}$$
$$=\frac{1}{4}.\frac{1}{z}\frac{1}{\frac{1}{z}-(-1)}-\frac{1}{4}.\frac{1}{3-(-z)}$$
$$=\frac{1}{4}.\frac{1}{z}\frac{1}{1-(-\frac{1}{z})}-\frac{1}{4}.\frac{1}{3}.\frac{1}{1-(-\frac{z}{3})}$$
and now both fraction can be expanded using the geometric series
$$=\frac{1}{4}.\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}...)-\frac{1}{12}.(1-\frac{z}{3}+\frac{z^2}{3^2}...)$$
Is the expansion correct?
NOTE: What is written above is the entire question.