I just cannot seem to wrap my head around this problem and would really appreciate some guidance. So I know that a primitive $n^{th}$ root of unity is a complex number $z$ such that $z^n = 1$ but $z^m \neq 1$ for $0<m<n$. I also know that for $w = cos(\frac{2\pi}{n}) + isin(\frac{2\pi}{n})$, the $n$ complex solutions of $z^n = 1$ are $1,w,w^2,...,w^{n-1}$. I need to show three things:

1.) $w$ is always a primitive $n^{th}$ root of unity.
2.) $w^k$ is a primitive $n^{th}$ root of unity iff $gcd(k,n) = 1$
3.) If $z$ is any primitive $n^{th}$ root of unity, then $1,z,z^2,...,z^{n-1}$ are distinct and comprise all the $n^{th}$ roots of unity.

I'm pretty new to this type of rigorous proof and I'm not even sure where to start.

  • $\begingroup$ So you have to show to start that $w^n=1$ and $w^k\ne 1$ if $0<k<n$. $\endgroup$ – xarles Oct 3 '18 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.