n-th roots of unity: vertices of a regular polygon

The formula

The n-th roots of unity are the $n$ complex solutions of $z^n = 1$:

$z_k = e^{2 i k \pi / n} = \cos\left(\dfrac{2 k \pi}{n}\right) + i \sin\left(\dfrac{2 k \pi}{n}\right)$

for $k = 0, 1, 2, \ldots, n-1$.

Geometric picture

All these roots have modulus 1, so they lie on the unit circle.

They are evenly spaced by an angle $\dfrac{2\pi}{n}$, forming the vertices of a regular polygon with $n$ sides inscribed in the circle.

Cases $n = 2, 3, 4$

$n = 2$: $z^2 = 1$ ⇒ $z = 1$ or $z = -1$. Two diametrically opposite points.

$n = 3$: $z^3 = 1$ ⇒ $z = 1$, $j = e^{2i\pi/3}$, $j^2 = e^{-2i\pi/3}$. Equilateral triangle.

$n = 4$: $z^4 = 1$ ⇒ $z = 1, i, -1, -i$. Inscribed square.

Magic properties

Zero sum (if $n \geq 2$): $\sum_{k=0}^{n-1} z_k = 0$

Follows from the geometric sum formula.

Product = $(-1)^{n-1}$.

n-th roots of any complex

To solve $z^n = w$ where $w = r e^{i\alpha}$:

$z_k = r^{1/n} \, e^{i(\alpha + 2 k \pi)/n}$

for $k = 0, 1, \ldots, n-1$.

The number $j$ to memorize

$j = e^{2 i \pi / 3} = -\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}$

Properties: $j^3 = 1$, $1 + j + j^2 = 0$, $\overline{j} = j^2 = 1/j$.

Common mistakes

1. Thinking there is only one root ($z^n = 1$ has exactly $n$ solutions in $\mathbb{C}$).
2. Forgetting the $2\pi$ in the formula (not $\pi$).
3. Confusing $j$ (complex) with vector $\vec{j}$ (mind the conventions).

?À toi de jouer

Number of cube roots of 1 in ℂ?

A.3B.1C.2D.6

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