0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.