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.