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**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.