I want to find the Laurent series inside $1<|z|<3$ for $$f(z)=\dfrac{9z-1}{(z^2-9)(z^3+1)}$$
I asked a similar question earlier but now I have to combine it together and I'm not too sure how.
So for my partial fractions I got $\dfrac{-1}{6(z+3)}$ , $\dfrac{1}{6(z-3)}$ , $\dfrac{1}{(z^3+1)}$ So what I've done is to take partial fractions and ended up with the answer
$$f(z)=(-1/18)\sum_{n=0}^\infty (-1)^n(z/3)^n-(1/18)\sum_{n=0}^\infty (z/3)^n+\sum_{n=0}^\infty (-1)^n z^(n-2)$$
I'm not sure if this is right though