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.

  • $\begingroup$ If all you want is the radius of convergence around 0, say, it will be at the first singularity, i.e. smallest zero of the denominator. If you want the series, decompose $z^2+z+1$ using partial fractions and expand the result as geometric series. $\endgroup$ – sharding4 Jun 15 '17 at 3:10
  • $\begingroup$ sorry I forgot to add I want it around z=1 $\endgroup$ – allizdog Jun 15 '17 at 3:12
  • $\begingroup$ Same would apply as to the radius of convergence. The series will converge up to $(-1\pm\sqrt{3})/2$ $\endgroup$ – sharding4 Jun 15 '17 at 3:17

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

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

$A = \frac {1}{\phi - \phi'}$ 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 $\phi,\phi'$

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


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