7
$\begingroup$

Let $n$ be an integer, $n\ge2$. Let $m$ be a positive integer, $m\le n$, having no common factor with $n$. How can we select $m$ distinct complex $n$th roots of unity in such a way as to minimise the modulus of their sum?

The condition that $m$ and $n$ be coprime is included because if $m$ and $n$ have a common factor, it is easy to see that we can always find a sum which vanishes.

The obvious answer is that the roots chosen should be spread around the unit circle as evenly as possible. This does not appear to be correct: for example, if $n=9$, $m=4$ and if we write $\alpha=e^{2\pi i/9}$, then $$|1+\alpha^2+\alpha^4+\alpha^6|=2\cos\frac{2\pi}{9}-1=0.5321\ ,$$ while $$|1+\alpha+\alpha^4+\alpha^6|=2\cos\frac{\pi}{9}-2\cos\frac{2\pi}{9} =0.3473\ .$$

$\endgroup$
  • $\begingroup$ Is something known about the distribution of euclidean norm of the centroid of $m$-uples of $n$-th roots of unity? I am thinking that the minimum may be very difficult to find, but we may have something like "there is a choice that grants a distance less than $\frac{2}{\sqrt{m}}$" through the probabilistic method. $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 4:56
  • $\begingroup$ We can also use something like (assuming $p\mid m$) choosing $p$ roots every time, minimizing their sum, then choosing other $p$ roots among the remaining one, minimizing their sum... $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 5:01
  • 2
    $\begingroup$ There is the nice $m=5,n=12$ where $5=3+2$, and you choose a trio with sum=0, and a pair with sum=0. $\endgroup$ – Empy2 Aug 11 '14 at 5:46
  • $\begingroup$ @Michael good observation! That suggests that it may be possible to get a zero sum whenever $m$ is a "suitable" sum of small coprime factors of $n$. $\endgroup$ – David Aug 11 '14 at 6:09
  • 2
    $\begingroup$ All the cases with $m\ge5$ are hard (the cases with $m\le4$ are easy). You can see my paper, How Small Can a Sum of Roots of Unity Be?, The American Mathematical Monthly, Vol. 93, No. 6 (Jun. - Jul., 1986), pp. 457-459. See also mathoverflow.net/questions/46068/… $\endgroup$ – Gerry Myerson Aug 11 '14 at 6:12

Your Answer

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

Browse other questions tagged or ask your own question.