---
tag: 64742ASK9R
kind: proof
name: Proof of the description of the nth roots of unity
msc: [11R18]
proves: 057BH3JC9S
uses: [0S61N75H45, 03MTBVG3VE, 0BRXFNQ5JV, 04R3CY41VJ, 0WB57D9488, 0P4NQD746K, 0KW2X0TFBX, 0HYNZE940T, 05V7J4WD1C, 0YVRNEDSVG, 0ZZRSX91QR, 0SQSQP748R, 0CHQKZFK5S]
sources:
  - url: https://en.wikipedia.org/wiki/Root_of_unity
    locator: Trigonometric expression
---

## Proof

We use the polar form of $z$ and de Moivre's formula.

1. Let $n \ge 1$. For an integer $k$, let $\omega_k = \cos \frac{2 \pi k}{n} + i \sin \frac{2 \pi k}{n}$.
For real arguments the power series of $\cos$ and $\sin$ are the real [[sine and cosine|0WB57D9488]].
So [[Euler's formula|04R3CY41VJ]] gives $\omega_k = e^{2 \pi i k / n}$.
2. First let $0 \le k < n$. By [[de Moivre's formula|03MTBVG3VE]], $\omega_k^n = \cos(2 \pi k) + i \sin(2 \pi k)$.
3. Since $\sin$ and $\cos$ are [[$2\pi$-periodic|0P4NQD746K]], applied $k$ times, $\cos(2 \pi k) = \cos 0 = 1$ and $\sin(2 \pi k) = \sin 0 = 0$.
So $\omega_k^n = 1$, and $\omega_k$ is an $n$th [[root of unity|0S61N75H45]].
4. Conversely, let $z$ be a complex number with $z^n = 1$. Then $z \neq 0$, since $0^n = 0 \neq 1$.
5. Let $r = |z| > 0$ be the [[modulus|0KW2X0TFBX]]. Since the [[modulus is multiplicative|0HYNZE940T]], induction on $n$ gives $|z^n| = r^n$.
So $r^n = |1| = 1$.
6. Claim: $r = 1$. If $0 < r < 1$, then $0 < r^m < 1$ for all $m \ge 1$, by induction: $r^{m+1} = r \cdot r^m < r < 1$.
If $r > 1$, then in the same way $r^m > 1$ for all $m \ge 1$. Both contradict $r^n = 1$.
7. Let $\theta = \operatorname{Arg} z$. By the [[polar form|0BRXFNQ5JV]] and step 6, $z = \cos \theta + i \sin \theta$.
8. By de Moivre's formula, $1 = z^n = \cos(n \theta) + i \sin(n \theta)$.
By [[equality of complex numbers|05V7J4WD1C]], $\cos(n \theta) = 1$ and $\sin(n \theta) = 0$.
9. Let $m = \floor{(\pi - n \theta) / (2 \pi)}$, using the [[floor function|0YVRNEDSVG]], and let $t = n \theta + 2 \pi m$.
Then $2 \pi m \le \pi - n \theta < 2 \pi m + 2 \pi$, so $-\pi < t \le \pi$.
10. By periodicity, applied $\abs{m}$ times, $\cos t = 1$ and $\sin t = 0$. So $t$ is an [[argument|0ZZRSX91QR]] of $1$ in $(-\pi, \pi]$.
11. Also $0$ is an argument of $1$ in $(-\pi, \pi]$. By uniqueness of the [[principal value of the argument|0SQSQP748R]], $t = 0$.
12. Let $j = -m$. Then $n \theta = 2 \pi j$, so $\theta = 2 \pi j / n$ with $j$ an integer.
13. By the [[division algorithm|0CHQKZFK5S]], $j = n q + k$ with integers $q$ and $k$ and $0 \le k < n$.
Then $\theta = \frac{2 \pi k}{n} + 2 \pi q$.
14. By periodicity, applied $\abs{q}$ times, $\cos \theta = \cos \frac{2 \pi k}{n}$ and $\sin \theta = \sin \frac{2 \pi k}{n}$.
So $z = \omega_k$ with $0 \le k < n$.
15. By steps 1 to 3 and steps 4 to 14, $z^n = 1$ holds if and only if $z = e^{2 \pi i k / n}$ for some integer $k$ with $0 \le k < n$.

## Idea

An $n$th root of unity has modulus $1$, so it is $\cos \theta + i \sin \theta$ for some angle $\theta$.
By de Moivre's formula, its $n$th power has angle $n \theta$, and this power is $1$ exactly when $n \theta$ is a multiple of $2 \pi$.
Reducing the multiple modulo $n$ leaves the $n$ angles $2 \pi k / n$ with $0 \le k < n$.