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
 A: Without computation, one can reply that the radius is $1$ as that is the absolute value of the poles.
If you inssist onexpicit computation, remembre the geometric series $\sum_{n=0}^\infty q^n=\frac1{1-q}$.
Thus 
$$\frac 1{1-z^3}=1+z^3+z^6+z^9+\ldots$$
and after multiplying with $1-z^2$:
$$f(z)=1-z^2+z^3-z^5+z^6.z^8+z^9-z^{11}\pm\ldots$$
One readily shows that the coefficient $a_n$ of $z^n$ is $1$ or $0$ or $-1$, depending on wether $z\mod 3$ is $0$ or $1$ or $2$.
Thus $$\frac1R=\limsup\sqrt[n]{|a_n|}=1.$$
A: $$
\frac{z^2-1}{z^3-1} = \frac{(z-1)(z+1)}{(z-1)(z^2+z+1)} = \frac{z+1}{z^2+z+1}
$$
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.
