sum of unit roots is also unit root

Asked Jan 01 '14 at 18:08

Active Jan 01 '14 at 18:38

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A complex number $\epsilon$ is a unit root if $\epsilon^n=1$ for some positive integer $n$.

How would one prove that if $\epsilon_1, \epsilon_2,\dots, \epsilon_k$ are all unit roots, and $|\epsilon_1+\epsilon_2+\dots+\epsilon_k|=1$, then $\epsilon_1+\epsilon_2+\dots+\epsilon_k$ is also a unit root ?

I have no clue about it. Could anyone help me? Thanks a lot.

p.s. I only know the problem was from a math student who got perfect score on the IMO

edited Jan 01 '14 at 18:37

Michael Hardy

asked Jan 01 '14 at 18:08

ziang chen

7,771

does $\epsilon_i$ include $1$ – Suraj M S Jan 01 '14 at 18:22
Did you mean that $\forall k\ \ n_k=n=\mathrm{const}$ where $\epsilon_k^{n_k}=1$? Or $\exists k_1\ne k_2\ \ n_{k_1}\ne n_{k_2}$? – Constructor Jan 01 '14 at 18:36
@Constructor it seems no difference: $\epsilon_i^{n_i}=1$ then $\epsilon_i^{n_1n_2\dotsb n_k}=1$ – ziang chen Jan 01 '14 at 18:41
@ziangchen You are right, of course. Thank you. – Constructor Jan 01 '14 at 18:44
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See http://mathoverflow.net/questions/126966/can-the-sum-of-two-roots-of-unity-be-a-root-of-unity. – Dietrich Burde Jan 01 '14 at 19:05
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I think this was answered here. – Nick Strehlke Jan 01 '14 at 19:06

sum of unit roots is also unit root

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