The nth Roots of Unity

Definition. For a positive integer $n$, the solutions of $z^n=1$ are the $n$th roots of unity. In polar form, $z^n=1$ is written

Applying the root formula from §2.8.2 (with $r=1,\theta=0$), the $n$th roots of unity are

Definition. A complex number $z$ is called an $n$th root of unity if and only if $z^n=1$. Denote $\omega=e^{2\pi i/n}=\cos\dfrac{2\pi}n+i\sin\dfrac{2\pi}n$ (the value at $k=1$); then $\omega^n=\left(e^{2\pi i/n}\right)^n=e^{2\pi i}=1$, so $\omega$ is itself an $n$th root of unity, and (by the root formula) the full list of $n$th roots of unity is exactly

These $n$ complex numbers are the points/vertices of a regular polygon of $n$ sides inscribed in the unit circle (since, as in §2.8.2, all $n$th roots of unity have modulus $1$ and are equally spaced by $\dfrac{2\pi}n$).

The $n$th roots of unity $1,\omega,\omega^2,\dots,\omega^{n-1}$ form a geometric progression with common ratio $\omega$.

• Sum: $1+\omega+\omega^2+\cdots+\omega^{n-1}=\dfrac{\omega^n-1}{\omega-1}=0$ (since $\omega^n=1$ and $\omega\ne1$, being $n\ge2$).
• Product: $1\cdot\omega\cdot\omega^2\cdots\omega^{n-1}=\omega^{0+1+\cdots+(n-1)}=\omega^{n(n-1)/2}=\left(e^{2\pi i/n}\right)^{n(n-1)/2}=e^{\pi i(n-1)}=(-1)^{n-1}.$

Note — summary facts about the $n$th roots of unity:

1. All $n$ roots lie in Geometric Progression.
2. The sum of the $n$ roots is always $0$.
3. The product of the $n$ roots is $(-1)^{n-1}$.
4. All $n$ roots lie on a circle of radius $1$ centred at the origin, dividing it into $n$ equal parts and forming a regular $n$-gon.

Cube roots of unity ($n=3$). Solving $z^3=1$ by the same method: $z=\cos\dfrac{2k\pi}3+i\sin\dfrac{2k\pi}3$ for $k=0,1,2$, giving

with $\omega^3=1$ and $1+\omega+\omega^2=0$ (matching the general sum result at $n=3$) — the two identities used constantly to reduce any expression in $\omega$ (e.g. $\omega^4=\omega\cdot\omega^3=\omega$, and $1+\omega=-\omega^2$).

Fourth roots of unity ($n=4$). Similarly, solving $z^4=1$ gives $1,\ i,\ -1,\ -i$. …