Find the Laurent series of $1/(z^3+1)$ in the annulus $1<|z|<3$

Question

I can't figure it out how to solve this problem:

Find the Laurent Series of the function $$f(z)=\frac{1}{(z^3+1)}$$ valid in $A=\{z \in \mathbb{C} : 1 < |z|<3\}$

My thoughts is to get the roots of unity at these points? Then make a laurent series for each point

Answer 1

HINT:

Note that

$$ f(z) = \frac{1}{z^3}\frac{1}{1+1/z^3} $$

and remember that

$$ \frac{1}{1+\omega} = \sum_{k=0}^{+\infty}(-1)^k \omega^k ~~~\mbox{for}~~~ |\omega| < 1 $$

Answer 2

Yes, you are on the right track, despite the easier method mentioned in the hint. Let the roots of $z^3+1$ be $z_0,z_1$ and $z_3$ You could then write you function as $\frac{1}{(z-z_0)(z-z_1)(z-z_2)}$, then you could expand $f(z)$ as partial fractions into three expressions and find the Laurent expansion of each of them about z=0.