Laurent series for $f(z)=\frac1{z^3 +1}$ on $0<|z-1|<\sqrt3$ I have this complex function $$f(z)=\dfrac{1}{z^3 +1}$$ and I have to find the Laurent series for $$0<|z+1|<\sqrt{3}$$
First, I consider that $$f(z) = \dfrac{1}{(z+1)(z^2 - z +1)} = \dfrac{2-z}{3(z^2 - z+1)} + \dfrac{1}{3(z+1)}$$
The second term is ok but I don't know how to expand the second term. Can anyone help me? Thanks a lot.

Let $a$ and $b$ be the complex cubic roots of $-1$, so $z^3+1=(z+1)(z-a)(z-b)$. Now you can find the partial fraction decomposition
$$
\frac{1}{z^3+1}=\frac{A}{z+1}+\frac{B}{z-a}+\frac{C}{z-b}
$$
Then
$$
\frac{1}{z-c}=\frac{1}{z+1-(c+1)}=-\frac{1}{c+1}\frac{1}{1-\dfrac{z+1}{c+1}}
$$
and then
$$
\frac{1}{z-c}=-\frac{1}{c+1}\frac{1}{1-\dfrac{z+1}{c+1}}=
-\frac{1}{c+1}\sum_{k\ge0}\frac{(z+1)^k}{(c+1)^k}=
-\sum_{k\ge0}\frac{(z+1)^k}{(c+1)^{k+1}}
$$
Do this for $c=a$ and $c=b$, then multiply by the coefficients $B$ and $C$ you found.
