Finding the radius of convergence of Laurent series

Question

I have this problem I want to solve: Find the radius of convergence of the power series of the function $\frac{1}{z^2+z+1}$ around $z=1$. I tried obtaining the derivatives but I could not simplify anything. I thought maybe I have to use the geometric series but I don't know how.

Answer 1

$(z^2 + z + 1) = (z-\phi)(z-\phi')$

$\frac {1}{z^2 + z + 1} = \frac {1}{(z-\phi)(z-\phi')} = A(\frac {1}{z-\phi} - \frac {1}{z-\phi'})$

$A = \frac {1}{\phi - \phi'}$ Not that it really matters in the radius of convergence.

$\frac {1}{z-\phi} = \frac {1}{(z-1) + 1 - \phi} = (\frac 1{\phi-1})\left(\frac {1}{1 - \frac {z-1}{\phi-1}}\right) = \frac 1{\phi-1} \sum_\limits{n=0}^\infty \left(\frac {z-1}{\phi-1}\right)^n$

and that series converges when $|z-1|<|\phi-1|$

But that is a Taylor series and not a Laurent series.

$\frac {1}{z-\phi} = (\frac 1{z-1})\left(\frac {1}{1 - \frac {\phi-1}{z-1}}\right) = \frac {1}{\phi-1}\sum_\limits{n=1}^\infty \left(\frac {\phi-1}{z-1}\right)^n$

is a Laurent series.

Which converges when $|z-1| > |\phi - 1|$

All that is left is to find $\phi,\phi'$

$\phi = \frac 12 + \frac {\sqrt 3}2i$
$\phi' = \frac 12 - \frac {\sqrt 3}2i$