The $n^{th}$ root of $1$

Question

How do I prove that the $n^{th}$ root of $1$ forms the vertices of a unit $n$-gon in the complex plane? For example when I graph the $3$ solutions of $\sqrt[3]{1}$ they form the vertices of a unit triangle-

I would also like to show that the $n$-gon formed by $\sqrt[n]{1}$ has its center at the co-ordinate $(0,0)$ and one of its vertices in the point $(1,0)$. The second part is simple: $1^n=1$, so one of the points must be at $(1,0)$. I am having trouble proving the rest. Thanks in advance.

P.S. I don't know how to tag this question, so feel free to edit the tags.

Answer 1

Hint:

$$|\sqrt[n]{z}| = \sqrt[n]{|z|}$$

$$\arg(\sqrt[n]{z}) = \frac{\arg{z} + 2k\pi}{n} $$

With $k$ varying from $0$ to $n-1$. You'd see that the $n$-th roots of the unity equidist from the origin and are equally spaced: thus, they are the vertices of a regular $n$-gon

Answer 2

Let K be the point $k=\cos\theta+i\sin\theta$, O the point 0, and A the point 1. Then the angle AOK is $\theta$. The point $k^n$ is $\cos n\theta+i\sin n\theta$, so $k$ is a root of unity if $\theta=\phi,2\phi,3\phi,\dots$ or $n\phi$, where $\phi=2\pi/n$. If we call the corresponding points $K_1,K_2,\dots$, then angle $K_rOK_{r+1}$ is $\phi$. You should now be able to prove that the $K_i$ form a regular $n$-gon.

Answer 3

Hint:Use De Moivre's formula to see the polar representation of the solutions