Find the Laurent series expansion for: $\frac{1}{z^2-3z-4}$

Question

Find the Laurent series expansion for: $$\frac{1}{z^2-3z-4}$$ which converges for$\:$ $1 < |z| < 3.$

I understand the concept of Laurents expansion but I cant seem to make progress in this question so any help will be appreciated.

Answer 1

$$\frac{1}{z^2-3z-4} = \frac{1}{(z+1)(z-4)}= \frac{1}{5}\left(\frac{1}{z-4}-\frac{1}{z+1}\right)$$

We look at the geometric series. For the first term we have $$\frac{1}{z-4}=\frac{-1/4}{1-z/4}=\sum_{n=0}^{\infty }\left( -\frac{1}{4} \right) \left( \frac{z}{4}\right) ^{n}\qquad \left\vert z\right\vert <4\implies|z|<3 \\ =\sum_{n=0}^\infty \frac{-1}{4^{n+1}}z^n$$ For the second term we have the well known $$\frac{1}{1+z}=\frac{1/z}{1-(-1/z)}=\sum_{n=0}^{\infty }\left(\frac{1}{z}\right)\left( \frac{-1}{z}\right) ^{n}\qquad \left\vert z\right\vert >1 \\ = \sum_{n=0}^\infty \frac{(-1)^n}{z^{n+1}}$$ We can combine these to get $$\sum_{n=0}^\infty \left(\frac{(-1)^n}{z^{n+1}}-\frac{1}{4^{n+1}}z^n\right) \qquad 1<|z|<3$$