nth Roots of Unity

This lesson covers finding the nth roots of complex numbers, with special attention to the roots of unity — the solutions of $z^n = 1$zn=1. This is one of the most elegant applications of De Moivre's Theorem.

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The nth Roots of a Complex Number

To solve $z^n = w$zn=w where $w = R(\cos\phi + i\sin\phi)$w=R(cosϕ+isinϕ):

Write $z = r(\cos\theta + i\sin\theta)$z=r(cosθ+isinθ). By De Moivre's Theorem:

$z^n = r^n(\cos(n\theta) + i\sin(n\theta)) = R(\cos\phi + i\sin\phi)$zn=rn(cos(nθ)+isin(nθ))=R(cosϕ+isinϕ)

Comparing moduli: $r^n = R$rn=R, so $r = R^{1/n}$r=R1/n (the positive real nth root).

Comparing arguments: $n\theta = \phi + 2k\pi$nθ=ϕ+2kπ for integer $k$k, so:

$\theta = \frac{\phi + 2k\pi}{n}$θ=nϕ+2kπ​

Taking $k = 0, 1, 2, \ldots, n - 1$k=0,1,2,…,n−1 gives exactly n distinct roots.

$z_k = R^{1/n}\left(\cos\frac{\phi + 2k\pi}{n} + i\sin\frac{\phi + 2k\pi}{n}\right), \quad k = 0, 1, \ldots, n - 1$zk​=R1/n(cosnϕ+2kπ​+isinnϕ+2kπ​),k=0,1,…,n−1

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The nth Roots of Unity

The nth roots of unity are the solutions of $z^n = 1$zn=1.

Since $1 = \cos 0 + i\sin 0$1=cos0+isin0 (so $R = 1$R=1, $\phi = 0$ϕ=0):

$z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \quad k = 0, 1, \ldots, n - 1$zk​=cosn2kπ​+isinn2kπ​,k=0,1,…,n−1

We often write $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$ω=cosn2π​+isinn2π​ (the primitive nth root of unity), and then:

$z_k = \omega^k, \quad k = 0, 1, \ldots, n - 1$zk​=ωk,k=0,1,…,n−1

The $n$n roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$1,ω,ω2,…,ωn−1.

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Geometric Properties

The $n$n roots of unity:

• All lie on the unit circle ($|z| = 1$∣z∣=1)
• Are equally spaced around the circle, separated by angles of $\frac{2\pi}{n}$n2π​
• Form the vertices of a regular $n$n-gon inscribed in the unit circle
• One root is always $z = 1$z=1

Sum of the Roots

The sum of all $n$nth roots of unity is:

$1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$1+ω+ω2+⋯+ωn−1=0

This is a geometric series with first term 1, ratio $\omega$ω, and $n$n terms:

$\sum_{k=0}^{n-1} \omega^k = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0 \quad (\text{since } \omega \neq 1)$∑k=0n−1​ωk=1−ω1−ωn​=1−ω1−1​=0(since ω=1)

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Worked Examples

Example 1: Cube Roots of Unity

Solve $z^3 = 1$z3=1.

$z_k = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}, \quad k = 0, 1, 2$zk​=cos32kπ​+isin32kπ​,k=0,1,2

$z_0 = 1$z0​=1 $z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$z1​=cos32π​+isin32π​=−21​+23​​i $z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$z2​=cos34π​+isin34π​=−21​−23​​i

The three roots form an equilateral triangle inscribed in the unit circle.

Note: $z_2 = \bar{z}_1$z2​=zˉ1​ — the complex roots come as a conjugate pair.

We write $\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ω=−21​+23​​i, and the roots are $1, \omega, \omega^2$1,ω,ω2.

Properties: $\omega^3 = 1$ω3=1, $1 + \omega + \omega^2 = 0$1+ω+ω2=0, $\omega^2 = \bar{\omega}$ω2=ωˉ.

Example 2: Fourth Roots of Unity

Solve $z^4 = 1$z4=1.

$z_k = \cos\frac{k\pi}{2} + i\sin\frac{k\pi}{2}, \quad k = 0, 1, 2, 3$zk​=cos2kπ​+isin2kπ​,k=0,1,2,3

The roots are: $1, i, -1, -i$1,i,−1,−i.

These form a square inscribed in the unit circle with vertices on the axes.

Example 3: Fifth Roots of Unity

Solve $z^5 = 1$z5=1.

$z_k = \cos\frac{2k\pi}{5} + i\sin\frac{2k\pi}{5}, \quad k = 0, 1, 2, 3, 4$zk​=cos52kπ​+isin52kπ​,k=0,1,2,3,4

These five roots form a regular pentagon inscribed in the unit circle.

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nth Roots of a General Complex Number