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I can see this is true for the sum of two roots of unity with some basic trigonometry (the resulting argument is the half the sum of the original arguments, and so must also be a rational multiple of Pi) but trying to extend this method to the sum of more than two doesn't seem to work.

Is there some other way of showing this? It seems related to this question: Sums of roots of unity but I'm not sure if that can be extended to this.

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  • For the sum to be a root of unity of the same order see http://mathoverflow.net/questions/126966/can-the-sum-of-two-roots-of-unity-be-a-root-of-unity. – Dietrich Burde Oct 15 '13 at 13:21

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I've answered my own question! I hope that's okay.

I found a counterexample using trig. The sum of the three roots of unity with arguments $\frac{11}{6} \pi, \frac{1}{3} \pi, \frac{1}{3} \pi$ has argument $\arctan(\frac{1}{3}) + \frac{1}{12} \pi$.

By this answer, this cannot be a rational multiple of Pi.

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  • This property is unchanged by a rotation of the complex plane, so you can assume that one of these two roots of unity is 1. In particular, consider $1 + i + i$. The argument of this number is $\arctan(2)$, and whether $\arctan(2)$ is a rational multiple of $\pi$ is precisely the question (of the answer) that you link to (and of course, the answer is that is is not...so $1 + i + i$ is also a counter exmaple). – Tyson Williams Apr 16 '14 at 20:08