Expand the function \(f(z)=\frac{1}{(z+1)(z+3)}\) in a Laurent series, which is valid for \(1<|z|<3\).

Use partial fractions and a binomial series expansion, \((1+x)^{-1}=1-x+x^{2}-\) \(x^{3} \ldots\), for \(x\) values in the interval of convergence of \(-1