It's known that sum of all $n$'th roots of some $z \in \mathbb C$ with $|z| = 1$ is zero (if $n \geqslant 2$).

Is it true that any zero sum of roots of unity can be decomposed in this way? That is if we have $\zeta_1 + \ldots + \zeta_s = 0,$ where $\zeta_i^n = 1,$ then we can partition it in groups $\{\xi_1, \ldots, \xi_k\}$ of all $k$'th roots of some $z$ with $|z| = 1$ (with zero sum of elements in this group), with possibly different $k$ for different groups?


Example: if $\zeta_1 + \zeta_2 = 0,$ then $\zeta_1^2 = \zeta_2^2 = z,$ so $\zeta_1, \zeta_2$ are square roots of this $z.$ Is it not hard to show that if $\zeta_1 + \zeta_2 + \zeta_3 = 0,$ then $\zeta_1^3 = \zeta_2^3 = \zeta_3^3 = z,$ so $\zeta_1, \zeta_2, \zeta_3$ are cubic roots of some $z$.

If $\zeta_1 + \zeta_2 + \zeta_3 + \zeta_4 = 0,$ it seems there are two possiblities - like $\{1, i, -1, -i\}$ (one part), or like $\{e^{\pi i/3}, e^{4\pi i/3}\}, \{e^{2\pi i/3},e^{10\pi i/3}\}.$ But I don't know what is the argumentation for the general case.

  • $\begingroup$ Consider the 8-th root of unity $r_k=e^{k 2\pi i/8}$ for $k=0\cdots 7$. The sum of roots at $k=1,3,5,7$ is zero but this is not all the roots. Did I misunderstand the question? $\endgroup$
    – Maesumi
    Aug 30, 2013 at 18:14
  • $\begingroup$ ) Yes. They all are 4-th roots of $e^{2\pi i}{2} = i$ The question is: is it possible to decompose any zero sum of unity's roots in symmetric ones or not? $\endgroup$
    – ptashek
    Aug 30, 2013 at 19:12
  • $\begingroup$ The $r_k$ in Maesumi's comment are $4$th roots of $-1$, not of $i$. $\endgroup$
    – anon
    Aug 31, 2013 at 15:59
  • $\begingroup$ Yes. There are some typos in the comment. $\endgroup$
    – ptashek
    Aug 31, 2013 at 16:22

1 Answer 1


No, for example pick $\zeta = \exp \frac{i\pi}{15}$, a primitive $30$th root of unity.

Then $0 = (1 + \zeta^{10} + \zeta^{20}) + \zeta^{15}(1 + \zeta^6 + \zeta^{12} + \zeta^{18} + \zeta^{24}) - (1 + \zeta^{15}) \\ = \zeta^3 + \zeta^9 + \zeta^{10} + \zeta^{20} + \zeta^{21} + \zeta^{27}$

But you can't partition those $6$ roots into vertices of regular $k$-gons.

However, it is true that you can always add vertices of regular $k$-gons to your set of roots (possibly adding the same root multiple times) to obtain a new multiset that you can then arrange into a sum of vertices of regular $k$-gons (possibly using the same $k$-gon multiple times).

Let $\zeta_n = \exp{\frac{2i\pi}n}$ and $f$ be the map $\Bbb Z^n \to \Bbb Z[\zeta_n]$ given by $(a_i) \mapsto \sum a_i \zeta_n^i$.
The minimal polynomial of $\zeta_n$ over $\Bbb Q$ is the cyclotomic polynomial $\Phi_n$, of degree $\varphi(n)$. Hence $\Bbb Z[\zeta_n]$ is free of rank $\varphi(n)$.

If $n = \prod p_i^{d_i}$, then $\zeta$ is a primitive $n$th root of unity iff it is of order exactly $n$, and not $n/p$ for any $p$, iff $\zeta^{n/p}$ is a primitive $p$th root, for all $p$. Hence $\Phi_n$ is the gcd of the $\Phi_p(X^{n/p}) = 1 + X^{n/p} + X^{2n/p} + \ldots + X^{(p-1)n/p}$, and $\ker f$ is generated by those relations, which correspond to vertices of regular $p$-gons.

  • $\begingroup$ Thank you! But about your note about decomposition with possible adding regular $k$-gons - I have the dimension argument which says that its not true: consider module $A$ generated by all 14-th roots of unity. You say that all relations in is are generated by "regular" sums, representing regular $k$-gons. But we have only $7 + 2 + 1$ such sums (some of which are linearly depended). $\endgroup$
    – ptashek
    Sep 3, 2013 at 17:24
  • $\begingroup$ Its not enougth. Consider the map $f\colon \mathbb Z^{14} \to A.$ $A$ has rank 2, $\mathbb Z^{14}$ has rank 14, so $\ker f$ must have rank $12,$ and regular sums span only module of rank 10. So there are some non-regular zero sums. $\endgroup$
    – ptashek
    Sep 3, 2013 at 17:27
  • $\begingroup$ So, if my conclusions are correct, we have another, non-constructive proof of this negative result. $\endgroup$
    – ptashek
    Sep 3, 2013 at 17:29
  • 1
    $\begingroup$ @ptashek : $A$ has rank $\varphi(14)=6$ and not $2$. There is no counterexample with $14$th roots. You need at least $3$ distinct prime factors in $n$ if you want to find a counterexample with $n$th roots, so my example was the smallest one. $\endgroup$
    – mercio
    Sep 4, 2013 at 6:37
  • $\begingroup$ Thank you again, my intuition for some reason tells me that $A$ is discrete in $\mathbb C = \mathbb R^2$) $\endgroup$
    – ptashek
    Sep 4, 2013 at 14:35

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