MHB Roots of equation and probability

[anemone]

Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$. Find the probability that $\sqrt{2+\sqrt{3}}\le |v+w|$.

 

[lfdahl]

Spoiler: Suggested solution

We are given $1997$ equidistant points on the unit circle in the complex plane each of which is a solution to the equation $z^{1997}-1 = 0$. The angular spacing between neighbouring solutions is:

\[\Delta \theta = \frac{2\pi}{1997} \approx 0,0031463\]

WLOG we can let $u = 1 +i0$, and choose $w$ arbitrarily among the 1996 other possibilities. The angle, $\theta$, between $u$ and $w$, determines the absolute value of the sum $u + w$:

\[|u+w| = 2 \cos \frac{\theta }{2} \geq \sqrt{2+\sqrt{3}}\]

The low limit can be expressed as a cosine:

\[\sqrt{2+\sqrt{3}} = \sqrt{2}\sqrt{1+\frac{\sqrt{3}}{2}} = \sqrt{2}\sqrt{\cos 0 + \cos \frac{\pi}{6}} = \sqrt{2}\sqrt{2 \cos^2 \frac{\pi}{12}} = 2 \cos \frac{\pi}{12}\]

Hence, we have the condition: $\cos \frac{\theta}{2} \geq \cos \frac{\pi}{12}$ or $- \frac{\pi}{6}\leq \theta\leq \frac{\pi}{6}$

Thus, having chosen $u$, the $w$’s which fulfill the inequality may be chosen up to a max. angular distance of $\frac{\pi}{6}$ from $u$.

There are $\left \lfloor \frac{\pi}{6 \Delta \theta } \right \rfloor = 166$ solutions on both sides of $u$, which fulfill the inequality.

If $w$ is picked out at random, we get the probability

\[\frac{2\cdot 166}{1996} = \frac{83}{499} \approx 16,63 \%.\]

This result differs slightly from the fraction: $\frac{1}{6} = 16,666.. \%$, which would be obtained in the limit: $\Delta \theta \approx 0$ ($N \rightarrow \infty$).