Cube Roots of Unity

The root of unity is a number which is complex in nature and gives 1 if raised to the power of a positive integer n. These roots are used in different branches and topics of maths like number theory. It is also called as de Moivre system. Here we will discuss the cube roots of unity in detail.

What are Cube Roots of Unity?

Suppose the cube root of 1 is “a”,

i .e 3√1 = a.

According to the general cube roots definition, a3 = 1 or a3 – 1 = 0

(a3 – b3) = (a – b) ( a2 + 12 + 2 x a x 1)

  • (a3 – 13) = (a – 1)( a2 + 1 – 2a)
  • Either (a – 1) = 0 or ( a2 + 1 – 2a) = 0
  • a = 1 or ( a2 + 1 – 2a) = 0
  • a = -1 ± sqrt( 12 – 4 x 1 x1 ) / (2 x 1)
  • a = – 1 ± sqrt (- 3) / 2
  • a = -½ ± i sqrt (- 3) / 2

So, there are a total of three cube roots of unity as follows

  • a = 1, (Real numbers)
  • a = – ½ + i sqrt (- 3) / 2 (Conjugate complex numbers)
  • a = -½ – i sqrt (- 3) / 2 (Conjugate complex numbers)

These are imaginary cube roots of unity.

Properties of Cube roots of unity

1) One imaginary cube roots of unity is the square of the other.

\(\left ( \frac{-1+\sqrt{3i}}{2} \right )^{2}\) = ¼ [(-1)2 – 2 x 1 x √ 3 i + ( √ 3 i)2] = ¼(1 – 2√ 3i – 3) = (-1-√ 3 i) /2

And \(\left ( \frac{-1-\sqrt{3i}}{2} \right )^{2}\) = ¼ [(-1)2 + 2 x 1 x √ 3 i + ( √3 i)2] = ¼(1 + 2√ 3i – 3) = (-1+ √ 3 i) /2

2) If two imaginary cube roots are multiplied then the product we get is equal to 1.

Let’s take ω = (-1 + √3 i ) /2

ω2 = (-1 – √3 i ) /2

Now we will get the product of two imaginary cube roots as ω x ω2 = (-1 + √3 i ) / 2x (-1 – √3 i ) /2

Or ω3 = ¼[(-1)2 – (√3 i)2]

= ¼[( 1 – 3i2) = ¼ x 4 = 1.

3) As there are three cube roots of unity, their sum is zero, let’s see hoe

= 1 + (-1 + √3 i ) /2 + (-1 – √3 i ) /2

Or

1 + ω + ω2 = 1 – ½ + (√3 i ) /2 – ½ – (√3 i ) /2

= 0

Examples of Cube Roots of unity

Now you know that finding the cube root of unity is different from how to find the cube root of any number. Let’s practice it below.

Question: Let’s factorize the following: a2 + ab + b2

Solution: a2 – (-1)ab + 1.b2

a2 – (ω + ω2)ab + ω3 . b2

a2 – (ω + ω2)ab + ω3 . b2

a2 – abω -ab ω2 + ω3 . b2

a (a – bω) – bω2 (a – bω) = (a – bω2)(a – bω)

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