1
$\begingroup$

It is given a set $Z=\{z_i\}_{i=1 \dots n} \subset \mathbb{C}$, I'm interested in finding (the) minimal conditions under which $Z$ contains exactly the $n$-th roots of the unity, that is $Z=\{ e^{\frac{i2k\pi}{n}} \}_{k=0 \dots n-1}$. Although I don't have an answer yet, I made some attempts to see what properties may be useful for the task:

My first attempt

We require that (remember that $Z$ is a set of $n$ complex numbers):

(i) $|z|=1$ for every $z \in Z$.

(ii) $Z$ is closed under conjugation, that is if $z \in Z$, then also $\overline{z} \in Z$.

(iii) $\sum_{z \in Z} z =0$

Unfortunately, this is not enough: consider $Z=\{ e^{i(\frac{\pi}{4}+k\frac{\pi}{2})} \}_{k=0,1,2,3}$, then $Z$ satisfies the conditions, but of course those are not the $4$-th roots of the unity.

A further idea was to also require that $1 \in Z$, and this seems to work, if I'm not missing something. I'm iterested in different perspectives and approaches to this problem: more elegant or sinthetic conditions (provided mine are correct), famous theorems or results concerning this topic. Any contribution is appreciated.

  • 0
    How about $z^n_i -1 = 0$ and they all are different numbers?2017-02-08
  • 0
    That would be (almost) exactly the definition of the $n$-th roots of the unity, I'm looking for a different characterization...2017-02-08
  • 0
    Okay, let's see where it would lead :) am I right that statements like "only $n$-th roots of unity are also solution of another non-trivial equation/problem" do count as non-trivial characterization?2017-02-08
  • 0
    I would say yes :). To make things more clear, I was thinking about this problem as part of a "bigger" problem (maybe a competition question): at some point one finds a set of complex numbers which satisfies some properties, and this properties exactly tell (to be proved, one of the reasons of this post) that those are the $n$-th roots of unity, although they were "hidden".2017-02-08
  • 1
    Maybe replace (ii) with "closed under *multiplication*".2017-02-08
  • 0
    Sounds cool, will work on that as soon as I can! Thank you.2017-02-08

0 Answers 0