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I am having trouble figuring out the answer

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Around what point are you trying to compute a power series? $z=0$? – copper.hat Dec 12 '12 at 19:22
I'm going to assume, the problem does not specify. – Rachel Dec 12 '12 at 19:24

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 $$\frac1R=\limsup\sqrt[n]{|a_n|}=1.$$

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Assuming the expansion is about the origin. If it is about $z=11$ then the radius is $10$ :-). – copper.hat Dec 12 '12 at 19:24
@copper.hat Actually, it would be $>10$ since $z=1$ is no pole after all. :). – Hagen von Eitzen Dec 12 '12 at 19:28
@HagenvonEitzen, strictly speaking, $f$ is not defined at $z=1$ though it can be extended to be defined there. – lhf Dec 12 '12 at 19:43
@HagenvonEitzen: Oops, I made a removable mistake... It should be $\sqrt{(11+\frac{1}{2})^2+\frac{3}{4}}$. – copper.hat Dec 12 '12 at 19:45
@lhf: That $f$ is not defined at $z=1$ does not prevent its power series (expanded around $11$, say) from converging. – Hagen von Eitzen Dec 12 '12 at 20:40

$$ \frac{z^2-1}{z^3-1} = \frac{(z-1)(z+1)}{(z-1)(z^2+z+1)} = \frac{z+1}{z^2+z+1} $$ 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.

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