Find the radius of convergence of the Taylor series about \(z=1\) of \[ f(z)=\frac1{1+z^2+z^4+z^6+z^8+z^{10}}. \] Express the answer using only real numbers and square roots.
Solution
Let \[ Q(z)\coloneqq 1+z^2+z^4+z^6+z^8+z^{10}. \]
1. One has\[ (z^2-1)Q(z)=z^{12}-1. \]
This is the finite geometric-series identity \[ (w-1)(1+w+w^2+w^3+w^4+w^5)=w^6-1 \] with \(w=z^2\).
2. The zeros of \(Q\) are exactly the twelfth roots of unity other than\(1\) and \(-1\).
If \(Q(z)=0\), step 1 gives \[ z^{12}=1. \] Moreover, \[ Q(1)=Q(-1)=6, \] so neither \(1\) nor \(-1\) is a zero.
Conversely, if \[ z^{12}=1 \qquad\text{and}\qquad z\neq\pm1, \] then \(z^2-1\neq0\), and step 1 gives \[ Q(z)=0. \]
3. The poles of\[ f(z)=\frac1{Q(z)} \]nearest to \(z=1\) are\[ e^{i\pi/6} \qquad\text{and}\qquad e^{-i\pi/6}. \]
By step 2, the poles are \[ e^{ik\pi/6}, \qquad k=1,\ldots,5,7,\ldots,11. \] For any real \(\theta\), \[ \abs{1-e^{i\theta}}^2 = 2-2\cos\theta. \] Among the allowed nonzero angles modulo \(2\pi\), the smallest absolute angle from \(0\) is \(\pi/6\), attained exactly at \(\theta=\pm\pi/6\). Since \(2-2\cos\theta\) increases with \(\abs{\theta}\) on \([0,\pi]\), these are the nearest poles to \(1\).
4. Their distance from \(1\) is\[ \sqrt{2-\sqrt3}. \]
Using \[ \cos\frac{\pi}{6}=\frac{\sqrt3}{2}, \] step 3 gives \[ \begin{aligned} \abs{1-e^{i\pi/6}}^2 &= 2-2\cos\frac{\pi}{6}\\ &= 2-\sqrt3. \end{aligned} \] Taking the positive square root gives the stated distance.
5. The radius of convergence of the Taylor series of \(f\) about \(z=1\)is\[ \boxed{\sqrt{2-\sqrt3}}. \]
A Taylor series of a rational function about a point where it is analytic has radius equal to the distance from the center to the nearest pole. Step 4 computes that distance.
6. Q.E.D.
Step 5 gives the requested radius in terms of real numbers and square roots only.