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-Hi, everybody. This is a problem from an assignment dealing with, among other things, power series, liouville's theorem, the extended liouville theorem, the fundamental theorem of algebra, and analytic functions.

Sorry for not doing a proper post. Thanks for all the help!

We are asked to find the radius of convergence R of the power series expansion about z = 1 of the function

$$ f(z) = \frac{1}{1 + z^2 + z^4 + z^6 + z^8 + z^{10}} $$

in terms of real numbers and square roots only. (That's how the problem is phrased :-/).

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Hint: multiply top and bottom by $1-z^2$. – Gerry Myerson Oct 14 '12 at 8:52
By Gerry's hint, the poles are at primitive 12th roots of unity. The radius of convergence is the distance of the closest pole to 1. It can be given as a single real number, but in fact also as an expression using rational numbers and square roots via the half-angle formulas for $\sin$ and $\cos$. – Hagen von Eitzen Oct 14 '12 at 9:02
@Hagen: that hint actually shows that the poles are at all the 12th roots of unity apart from $\pm 1$. – Chris Eagle Oct 14 '12 at 9:05
@ChrisEagle: Oops, jes, of course - there are 10 poles after all :) – Hagen von Eitzen Oct 14 '12 at 9:12
The distance from $1$ to the 12th roots of unity is $$2 \sin \frac{k\pi}{12}$$ where $k=1,2, \cdots, 12$. – PAD Oct 14 '12 at 9:19
up vote 1 down vote accepted

A full answer to the question as the question is phrased at the moment is $$ R=\sqrt{2-\sqrt3}. $$ Nota: If the question is rephrased to become more in line with the way the site is supposed to function, then the present answer might be rephrased to become more in line with the way the site is supposed to function (until then, why should it be?).

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The denominator factors like $(z^2+1)(z^2+z+1)(z^2-z+1)(z^4-z^2+1)$. The singular points are the roots of this polynomial. The closest distance from $1$ to the 10 roots is the radius of convergence.

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