Prove the radius of convergence of the complex power series $(z^2 - 1)/(z^3-1)$ about $z =2$ is sqrt(7)

Question

So first I took the partial fractions and removed the singularity $z-1$ so I have $f(z) = \frac{z+1}{z^2+z+1}$. Then I took the complex root of the denominator so now I have $f(z) = \frac{z+1}{(z-e^{2/3\pi i})(z+e^{2/3\pi i})}$. I am confused on where to go from here. I have tried plugging 2 in for $z$, trying to find zero points by using the unit circle, I am lost.

Answer 1

let $w = z-2\\z = w+2$

$\frac {w+3}{w^2+4w + 4 +w + 2 + 1}\\ \frac {w+3}{w^2+5w + 7}$

Now we have a to find the radius of convergence of series centered at 0.

$\frac {w+3}{(w + \frac {5 + i\sqrt 3}{2})(w + \frac {5 - i\sqrt 3}{2})}\\ \frac {A}{(w + \frac {5 + i\sqrt 3}{2})} + \frac {B}{(w + \frac {5 - i\sqrt 3}{2})}\\ \frac {A}{\frac {5+i\sqrt 3}{2}}\sum \frac {w^n}{(\frac {-5 + i\sqrt 3}{2})^n} + \frac {B}{\frac {-5 - i\sqrt 3}{2}}\sum \frac {w^n}{(\frac {-5 - i\sqrt 3}{2})^n} $

And what is the radius of convergence of those series?

I have done a little bit more work than is absolutely necessary, to show the motivation....

As soon as you had factored the denominator, you could have said.

$|\frac {5 + i\sqrt 3}{2}| = \sqrt 7$