Cube Roots of Unity | Mathematics for NDA PDF Download

Cube Root of Unity

There are three cube roots of unity as the name suggests: 1, ω, ω². Here the roots ω and ω² are imaginary roots and one root is a square of the other root. The product of the imaginary roots of the cube root of unity is equal to 1(ω · ω² = ω³ = 1), and the sum of the cube roots of unity is equal to zero.(1 + ω + ω² = 0).

The cube root of unity is represented as ∛1 and it has three roots. The three cube roots of unity are 1, ω, ω², which on multiplication gives the answer of unity (1). Among the roots of the cube roots of unity, one root is a real root and the other two roots are imaginary roots. The values of the imaginary cube roots of unity are as follows.

• ω = (-1 + i√3) / 2
• ω² = (-1 - i√3) / 2

Verification of ω and ω²:
Why is (-1 - i√3) / 2 called ω²?
Let's square ω and see what we will get:
ω² = ω × ω
= [(-1 + i√3) / 2]²
= (-1 + i√3)² / 4
= (1 + i² (3) - 2i√3)/4
= (1 - 3 - 2i√3)/4
= (-2 - 2i√3)/4
= (-1 - i√3)/4


The symbol ω is referred to as omega. Thus, the imaginary cube roots of unity ω, ω² are read as omega and omega square respectively.

Product of Cube Roots of Unity

The product of the cube roots of unity is equal to 1. This can be understood from the below expression.
1 × ω × ω² = 1 × (-1 + i√3) / 2 × (-1 - i√3) / 2
= ((-1)² - (i√3)²) / (2×2)
= (1 - 3i²) / 4
= (1 - 3(-1)) / 4
= 4/4
=1
∴ 1 × ω × ω² = 1

Sum of Cube Roots of Unity

The sum of the cube roots of unity is equal to zero. This can be observed in the below expression.

1 + ω + ω² = 1 + (-1 + i√3) / 2 + (-1 - i√3) / 2 = 1 + (-1 + i√3 -1 - i√3) / 2, = 1 + (-2/2) = 1 - 1 = 0

∴ 1+ ω + ω² = 0

How To Find Cube Root Of Unity?

The cube root of unity can be represented as an expression ∛1 = a, and it has three roots. This expression can be further simplified using algebraic formulas to find the three roots of the cube roots of unity.
∛1 = a
1 = a³
a³ - 1 = 0
By the formula a³ - b³ = (a - b) (a² + ab + b²),
(a - 1)(a² + a + 1) = 0
a - 1 = 0 or a = 1
We will solve a² + a + 1 = 0 by quadratic formula.
a = (-1 ± √(1² - 4(1)(1)) / (2 · 1)
= (-1 ± √(-3)) / 2
= (-1 ± i√3) /2
∴ a = (-1 + i√3) / 2, or (-1 - i√3) / 2
From the above expression, the three cube roots of unity are 1, (-1 + i√3) / 2, and (-1 - i√3) / 2.

Deriving Cube Roots of Unity by De Moivre's Theorem

∛1 can be written as 1^1/3. In complex numbers, we have a theorem called "De Moivre's theorem" which is very useful in finding the real and complex roots of a real/complex number. This theorem says, (a + ib)ⁿ = (r (cos θ + i sin θ))ⁿ = rⁿ (cos nθ + i sin nθ), where n ∈ ℤ. Let us assume that a + ib = 1 = 1 + i(0). Then we have a = 1 and b = 0.
Then r = √(a² + b²) = √(1² + 0²) = 1 and θ = tan^-1|b/a| = tan^-1|0/1| = 0. We can write this angle as 0 + 2nπ, where n = 0, 1, and 2 (since we need to find three roots, 3 integer values are taken for n).
By substituting these values in De Moivre's theorem along with substituting n = 1/3:
(1 + i(0))^1/3 = 1 (cos 2nπ + i sin 2nπ)^1/3, where n = 0, 1, and 2.
⇒ 1^1/3 = cos 2nπ/3 + i sin 2nπ/3, where n = 0, 1, and 2.

• When n = 0, 1^1/3 = cos 0 + i sin 0 = 1 + i(0) = 1.
• When n = 1, 1^1/3 = cos 2π/3 + i sin 2π/3 = -1/2 + i (√3/2) = (-1 + i√3) / 2 (= ω)
• When n = 2, 1^1/3 = cos 4π/3 + i sin 4π/3 = -1/2 - i (√3/2) = (-1 - i√3) / 2 (= ω²)

Thus, the cube roots of unity by De Moivre's theorem are 1, (-1 + i√3) / 2, and (-1 - i√3) / 2.

Properties of Cube Root Of Unity

• The cube roots of unity has two imaginary roots (ω, ω²) and one real root (1).
• The sum of the roots of the cube root of unity is equal to zero.(1 + ω + ω² = 0)
• The square of one imaginary root (ω) of the cube root of unity is equal to another imaginary root(ω²).
• The product of the imaginary roots of the cube roots of unity is equal to 1.(ω × ω² = ω³ = 1)

Solved Examples