The Cube Roots of Unity

We will discuss here about the cube roots of unity and their properties.

Suppose let us assume that the cube root of 1 is z i.e., ∛1 = z.

Then, cubing both sides we get, z\(^{3}\) = 1

or, z\(^{3}\) - 1 = 0

or, (z - 1)(z\(^{2}\) + z + 1) = 0

Therefore, either z - 1 = 0 i.e., z = 1 or, z\(^{2}\) + z + 1 = 0

Therefore, z = \(\frac{-1\pm \sqrt{1^{2} - 4\cdot 1\cdot 1}}{2\cdot 1}\) = \(\frac{-1\pm \sqrt{- 3}}{2}\) = -\(\frac{1}{2}\) ± i\(\frac{√3}{2}\)

Therefore, the three cube roots of unity are

1, -\(\frac{1}{2}\) + i\(\frac{√3}{2}\) and -\(\frac{1}{2}\) - i\(\frac{√3}{2}\)

among them 1 is real number and the other two are conjugate complex numbers and they are also known as imaginary cube roots of unity.

Properties of the cube roots of unity:

Property I: Among the three cube roots of unity one of the cube roots is real and the other two are conjugate complex numbers.

The three cube roots of unity are 1, -\(\frac{1}{2}\) + i\(\frac{√3}{2}\) and -\(\frac{1}{2}\) - i\(\frac{√3}{2}\).

Hence, we conclude that from the cube roots of unity we get 1 is real and the other two i.e., \(\frac{1}{2}\) + i\(\frac{√3}{2}\) and -\(\frac{1}{2}\) - i\(\frac{√3}{2}\) are conjugate complex numbers.

Property II: Square of any one imaginary cube root of unity is equal to the other imaginary cube root of unity.

\((\frac{-1 + \sqrt{3}i}{2})^{2}\) = \(\frac{1}{4}\)[(- 1)^2 - 2 ∙ 1 ∙ √3i + (√3i)\(^{2}\)]

               = \(\frac{1}{4}\)[1 - 2√3i - 3]

               = \(\frac{-1 - \sqrt{3}i}{2}\),

And \((\frac{-1 - \sqrt{3}i}{2})^{2}\) = \(\frac{1}{4}\)[(1^2 + 2 ∙ 1 ∙ √3i + (√3i)\(^{2}\)]

                    = \(\frac{1}{4}\)[1 + 2√3 i - 3]

                    = \(\frac{-1 + \sqrt{3}i}{2}\),

Hence, we conclude that square of any cube root of unity is equal to the other.

Therefore, suppose ω\(^{2}\) is one imaginary cube root of unity then the other would be ω.

Property III: The product of the two imaginary cube roots is 1 or, the product of three cube roots of unity is 1.

Let us assume that, ω = \(\frac{-1 - \sqrt{3}i}{2}\); then, ω\(^{2}\) = \(\frac{-1 + \sqrt{3}i}{2}\)

Therefore, the product of the two imaginary or complex cube roots = ω ∙ ω\(^{2}\) = \(\frac{-1 - \sqrt{3}i}{2}\) × \(\frac{-1 + \sqrt{3}i}{2}\)

Or, ω\(^{3}\) = \(\frac{1}{4}\)[(-1)\(^{2}\) - (√3i)\(^{2}\)] = \(\frac{1}{4}\)[1 - 3i\(^{2}\)] = \(\frac{1}{4}\)[1 + 3] = \(\frac{1}{4}\) × 4 = 1.

Again, the cube roots of unity are 1, ω, ω\(^{2}\). So, product of cube roots of unity = 1 ∙ ω ∙ ω\(^{2}\) = ω\(^{3}\) = 1.

Therefore, product of the three cube roots of unity is 1.

Property IV: ω\(^{3}\) = 1

We know that ω is a root of the equation z\(^{3}\) - 1 = 0. Therefore, ω satisfies the equation z\(^{3}\) - 1 = 0. 

Consequently, ω\(^{3}\) - 1 = 0

or, ω = 1.

Note: Since ω\(^{3}\) = 1, hence, ω\(^{n}\) = ω\(^{m}\), where m is the least non-negative remainder obtained by dividing n by 3.

Property V: The sum of the three cube roots of unity is zero i.e., 1 + ω + ω\(^{2}\) = 0.

We know that, the sum of the three cube roots of unity = 1 + \(\frac{-1 - \sqrt{3}i}{2}\) + \(\frac{-1 + \sqrt{3}i}{2}\)

Or, 1 + ω + ω\(^{2}\) = 1 - \(\frac{1}{2}\) + \(\frac{√3}{2}\)i - \(\frac{1}{2}\) - \(\frac{√3}{2}\)i = 0.

Notes:

(i) The cube roots of 1 are 1, ω, ω\(^{2}\) where, ω = \(\frac{-1 - \sqrt{3}i}{2}\) or, \(\frac{-1 + \sqrt{3}i}{2}\)

(ii) 1 + ω + ω\(^{2}\) = 0 ⇒ 1 + ω = - ω\(^{2}\), 1 + ω\(^{2}\) = - ω and ω + ω\(^{2}\) = -1

(iii) ω\(^{4}\) = ω\(^{3}\) ∙ ω = 1 ∙ ω = ω;

ω\(^{5}\) = ω\(^{3}\) ∙ ω\(^{2}\) = 1 ∙ ω\(^{2}\) = ω\(^{2}\);

ω\(^{6}\) = (ω\(^{3}\))\(^{2}\) = (1)\(^{2}\) = 1.

In general, if n be a positive integer then,

ω\(^{3n}\) = (ω\(^{3}\))\(^{n}\) = 1\(^{n}\) = 1;

ω\(^{3n + 1}\) = ω\(^{3n}\) ∙ ω = 1 ∙ ω = ω;

ω\(^{3n + 2}\) = ω\(^{3n}\) ∙ ω\(^{2}\) = 1 ∙ ω\(^{2}\) = ω\(^{2}\).

Property VI: The reciprocal of each imaginary cube roots of unity is the other.

The imaginary cube roots of unity are ω and ω\(^{2}\), where ω = \(\frac{-1 + \sqrt{3}i}{2}\).

Therefore, ω ∙ ω\(^{2}\) = ω\(^{3}\) = 1

⇒ ω = \(\frac{1}{ω^{2}}\) and ω\(^{2}\) = \(\frac{1}{ω}\)

Hence, we conclude that the reciprocal of each imaginary cube roots of unity is the other.

Property VII: If ω and ω\(^{2}\) are the roots of the equation z\(^{2}\) + z + 1 = 0 then - ω and - ω\(^{2}\) are the roots of the equation  z\(^{2}\) - z + 1 = 0.

Property VIII: Cube roots of -1 are -1, - ω and - ω\(^{2}\).

11 and 12 Grade Math 

From The Cube Roots of Unity to HOME PAGE