Proving N-th Root of Unity: e^{\frac{2k\pi i}{n}}

[autre]

Homework Statement

Prove that $z=e^{\frac{2k\pi i}{n}},n\in\mathbb{N},k\in\mathbb{Z}, 0\leq k\leq n-1$ is an n-th root of unity.

The Attempt at a Solution

So I know I have to come to the conclusion that $z^{n}=1$. I'm thinking of using the property $e^{i\theta}=cos\theta+isin\theta$, but when I try to break it up like that I get something strange like $e^{\frac{2k\pi i}{n}}=cos(\frac{2k\pi}{n})+i( \frac{2k\pi}{n})$ and I have to get it into a form like $cos^{2}(\theta)+sin^{2}(\theta)$. Any ideas on how to do that/am I on the right track?

 

[SammyS]

Homework Statement

Prove that $z=e^{\frac{2k\pi i}{n}},n\in\mathbb{N},k\in\mathbb{Z}, 0\leq k\leq n-1$ is an n-th root of unity.

The Attempt at a Solution

So I know I have to come to the conclusion that $z^{n}=1$. I'm thinking of using the property $e^{i\theta}=cos\theta+isin\theta$, but when I try to break it up like that I get something strange like $e^{\frac{2k\pi i}{n}}=cos(\frac{2k\pi}{n})+i( \frac{2k\pi}{n})$ and I have to get it into a form like $cos^{2}(\theta)+sin^{2}(\theta)$. Any ideas on how to do that/am I on the right track?

Look at $\left(e^{\displaystyle \left(\frac{2k\pi i}{n}\right)}\right)^n\,.$

 

[autre]

Oh, so I get $\left(e^{{\displaystyle \left(\frac{2k\pi i}{n}\right)}}\right)^{n}=e^{2k\pi i}=cos(2k\pi)+isin(2k\pi)...$ which looks better. But how do I go from there?

 

[SammyS]

Oh, so I get $\left(e^{{\displaystyle \left(\frac{2k\pi i}{n}\right)}}\right)^{n}=e^{2k\pi i}=cos(2k\pi)+isin(2k\pi)...$ which looks better. But how do I go from there?

If k is an integer, that is equal to 1 .