How does Mathematica's RootOfUnityQ function work?

That is, how does Mathematica know if a number is a root of unity?


Let $x = \frac{1-i \sqrt{2+\sqrt{5}}}{1+i \sqrt{2+\sqrt{5}}}$.


In[1]:= RootOfUnityQ[x]
Out[1]= False

Wrong Answers:

This bahavior is not the same as

In[2]:= Reduce[Abs[x] == 1 && Simplify[Element[Arg[x]/(2 Pi), Rationals]]]
Out[2]= ArcTan[(2 Sqrt[2 + Sqrt[5]])/(1 + Sqrt[5])]/\[Pi] \[Element] Rationals

2 Answers 2


As long as we're giving methods that mimic the behavior, here is a quick implementation based on Kronecker's theorem.

RoUQ[u_] :=
 If[! (Abs[N[u]] == 1), False,
  If[! AlgebraicIntegerQ[u], False,
   (f = MinimalPolynomial[u, x];
    n = Exponent[f, x];
    cf = CoefficientList[f, x]/Coefficient[f, x^n]; 
    M = Table[
      If[j == i + 1, 1,
       If[i == n, -cf[[j]], 0]], {i, 1, n}, {j, 1, n}];
    lambda = First[Eigenvalues[M, 1]];
    Abs[N[lambda]] == 1)]] 

The sole reason for building the matrix $M$ is that Eigenvalues[M,1] is guaranteed to return the largest eigenvalue of $M$ (in absolute value), while I don't know how to tell FindRoot that it must give me the largest root.

J.M. below suggests just finding all the roots and taking the largest one. Based on a tiny amount of testing, I think that this is slower. Here is what I did:

(* Both methods will be tested on the same set of polynomials *)
data=Table[Sum[RandomInteger[100] x^j, {j,0,10}], {i,1,10^4}];

(* Find all roots and extract the largest *)
BigRoot1[f_]:=Max[Abs[x] /. NSolve[f==0,x]]

(* Find the largest eigenvalue of the companion matrix. *)
(n = Exponent[f, x];
    cf = CoefficientList[f, x]/Coefficient[f, x^n]; 
    M = Table[
      If[j == i + 1, 1,
       If[i == n, -cf[[j]], 0.0]], {i, 1, n}, {j, 1, n}];
    lambda = First[Eigenvalues[M, 1]];

Timing[Map[BigRoot1, data]][[1]]


Timing[Map[BigRoot2, data]][[1]]


I wouldn't take these results too seriously, because I'm sure both implementations do the list processing inefficiently, but it suggests that I'm not crazy to use Eigenvalues[].

  • $\begingroup$ Well, you could use x /. First @ Solve[p[x] == 0, x] to generate the roots, and then Pick[roots, #, Max[#]] &[Abs[roots]] to get the largest of them all in magnitude. But using the companion matrix is cool anyway. :) $\endgroup$ Commented Aug 15, 2012 at 23:37
  • $\begingroup$ @J.M. It strikes me as wasteful to compute all those roots which will be discarded. I've added some tests above to support this claim. $\endgroup$ Commented Aug 16, 2012 at 0:14
  • $\begingroup$ Hmm, right. Then, I suppose if one wants to use FindRoot[], there's always the option of trying to find the smallest root of $x^n p\left(\frac1{x}\right)$ and reciprocating that... the only reason I was somewhat lukewarm to using the companion matrix is that it looks uncomfortable to use $O(n^2)$ storage and $O(n^3)$ effort for just a single root... $\endgroup$ Commented Aug 16, 2012 at 1:45
  • $\begingroup$ @J.M. Yeah, memory usage is a good point. I guess the next question then is how often FindRoot[], starting at 0, fails to find the smallest root. A good first test would be p[x]=x^10-x^9+x^7-x^6+x^5-x^4+x^3-x+1, or perhaps this after the change of variable x /. E^(2 Pi I RandomReal[]) x. This has two roots at 0.85 and 1.176 and the other 8 are all on the unit circle with roughly equal spacing, so a random walk from 0 is very likely to hit one of the roots on the circle before the small root. I imagine FindRoot[] does something smarter than a random walk, though. $\endgroup$ Commented Aug 16, 2012 at 12:13

If Abs[x]==1 and Element[Arg[x]/(2 Pi), Rationals], then the number is a root of unity

$$ \arg[x]= 2 \pi \frac{p}{q}$$ $p,q\in \mathbb{Z}$

and $$ |x|=1 $$

$$ \to x^q=1 $$

$ |x^q|=|x|^q=1^q=1 $ and
$ \arg[x^q] = \arg[x] q = 2 \pi p $

So perhaps this mimics the behaviour

ru = TrueQ[Abs[#] == 1 && Simplify@Element[Arg[#]/(2 Pi), Rationals] ]&;
  • $\begingroup$ It does not mimic the behavior. I updated the question to discuss your answer. $\endgroup$ Commented Aug 1, 2012 at 19:48
  • $\begingroup$ @TysonWilliams in your counterexample, it's just a matter of adding TrueQ. There are probably other counterexamples where this doesn't work, but please post them. I'm editing my answer $\endgroup$
    – Rojo
    Commented Aug 1, 2012 at 20:08
  • $\begingroup$ @R.M I am not looking for (numerically) approximate answers. I want to know with certainty. $\endgroup$ Commented Aug 1, 2012 at 20:10
  • $\begingroup$ Ok, suppose that this is the definition of RootOfUnityQ. The purpose of my question was to be able to verify Mathematica's answer for myself. So my question now becoems this: how does Mathematica compute ArcTan[Sqrt[1/2 (1+Sqrt[5])]]/\[Pi]\[Element]Rationals? $\endgroup$ Commented Aug 1, 2012 at 21:34

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