正多边形和单位根

As a quick review, an $n$th root of unity is a number $z$ such that $z^n=1$.

Also, a regular polygon is a polygon with all sides congruent and all angles congruent.

Now, how are these two related? In fact, if we plot the $n$th roots of unity on the complex plane, they will form a regular $n$-sided polygon!

Suppose that we have a number $z=r\cdot cis\theta$. Let us say that $z$ is an $n$th root of unity. Thus, we have: $$z^n=1$$ Using De Moivre's Theorem, $$r^n\cdot cis(n\theta)=1$$ First, let us compare magnitudes: $$|z^n|=|1|$$ $$r^n=1$$ Since $r$ is a nonnegative real number, it follows that $r=1$.

Now, we compare angles: $$cis(n\theta)=1$$ $$cis(n\theta)=cis(0)$$ It follows that $n\theta$ is an integer multiple of $2\pi$. Thus, for an integer $k$, $$n\theta=2k\pi$$ $$\theta=\frac{2k\pi}{n}$$ Thus, we see that the roots are of the form $z=cis(\frac{2k\pi}{n})$ for $k=0,1,2,...,n-1$.

Call these points $A_0,A_1,...,A_{n-1}$, respectively. Let $O$ denote the origin.

Using the Law of Cosines on triangle $OA_0A_1$, $$A_0A_1^2=1^2+1^2-2\cdot 1\cdot 1\cos\frac{2\pi}{n}$$ The same holds for all of the other sides. Thus, $A_0A_1=A_1A_2=...=A_{n-1}A_0$.

Calculating $\angle A_0A_1A_2$, $$\angle A_0A_1A_2=2\angle A_0A_1O$$ $$\angle A_0A_1A_2=2\left(\frac{1}{2}(\pi-\frac{2\pi}{n})\right)$$ $$\angle A_0A_1A_2=\frac{(n-2)\pi}{n}$$ The same holds for all other interior angles of the polygon.

Thus, we have proven that the roots of unity form a regular polygon in the complex plane.