I have the following formula and I would like to know its Laurents expansion in the complex plane with the condition $3<|z|<\infty$ and $z \in \mathbb{C}$.
Formula: $f: \mathbb{C}\setminus\{-1,-3\} \rightarrow \mathbb{C}$ given by
$$f(z)=\frac{z}{(z+1)(z+3)} $$
This is what I tried:
$$f(z)=\frac{z}{(z+1)(z+3)}=\frac{-1}{2(1+z)}+ \frac{3}{2(z+3)}$$
If we just take a look at the first term, since the second term has an analogue calculation, we have the following:
$$k(z)=-\frac{1}{2(1+z)}= \sum_{n=-\infty}^{\infty} -\frac{z^n}{4\pi i} \int\frac{1}{m^{(n+1)}(1+m)}dm$$
The integrand has a pole with otder n+1 at m=0 and a pole at m=-1 with order 1. But since -1 is not contained in our domain, we can ignore this on. Now I should use again Cauchy Integral Formula, I guess? I do not how, since the order for my pole is n+1. So my problem is, is that I do not know how to compute this integral at $3<|z|< \infty$.
Hint
$$\frac{1}{z-a}= \frac{1}{z}\frac{1}{1-\frac{a}{z}}=\frac{1}{z} \left( \sum_{n=0}^\infty \frac{a^n}{z^n}\right)$$ when $|z| > |a|$.