I'm working on a problem, but I haven't been able to get much out of it...

Given that $\zeta$ is a primitive complex n-th root of unity, show that the complex numbers $1,\zeta,\zeta^2,...,\zeta^{n-1}$ are all distinct, meaning that no two of them are equal. Note that it is not specified which primitive n-th root $\zeta$ is.

I believe I have to incorporate the orders of these $\zeta$ elements in some way. From what I know, a primitive complex n-th root of unity is when $O(\zeta) = n$, where $O(\zeta)$ is the order of $\zeta$. With this in mind, how would I show each of the elements, $1,\zeta,\zeta^2,...,\zeta^{n-1}$ are all distinct..?

  • $\begingroup$ To show $1,\zeta,\zeta^2,...,\zeta^{n-1}$ are all distinct, suppose $$\zeta^a = \zeta^b$$ where $0 \le a < b \le n-1$. Divide both sides by $\zeta^a$, and a contradiction should be apparent. $\endgroup$ – quasi Mar 15 '17 at 22:34
  • $\begingroup$ Can you explain what the contradiction is? $\endgroup$ – user425349 Mar 15 '17 at 22:36
  • $\begingroup$ So $1 = \zeta^{b-a}$? $\endgroup$ – user425349 Mar 15 '17 at 22:38
  • $\begingroup$ Right, and $0 < b-a < n$, so therefore ... $\endgroup$ – quasi Mar 15 '17 at 22:38
  • 1
    $\begingroup$ Yes, all fixed. $\endgroup$ – quasi Mar 15 '17 at 22:49

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