Find the radius of convergence of a power series expansion of the rational function $f(z)=\frac{(z^2)-1}{(z^3)-1 }$ I am having trouble figuring out the answer

$$
\frac{z^2-1}{z^3-1} = \frac{(z-1)(z+1)}{(z-1)(z^2+z+1)} = \frac{z+1}{z^2+z+1}
$$
The denominator is $0$ when
$$z=\dfrac{-1\pm\sqrt{1^2-4\cdot1\cdot1}}{2} = \frac{-1\pm i\sqrt{3}}{2}. $$
If you want this in powers of $z$, i.e. powers of $(z-0)$, so the center is $0$, then the radius of convergence is the distance from the center, $0$, to the nearest point where the function fails to be well behaved, i.e from $0$ to either of $\dfrac{-1\pm i\sqrt{3}}{2}$.  They're both at the same distance:
$$
\left| \dfrac{-1\pm i\sqrt{3}}{2} \right| = \frac14+\frac34=1.
$$
So that is the radius of convergence.
