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Mar 22, 2022

Cube Roots of Unity

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Cube Roots of Unity

Cube Roots of Unity:

Let us denote the cube roots of unity be z.

Then, z = (1)1/3 ⇒ z3 – 1 = 0 ⇒ (z – 1) (z2 + z + 1) = 0

When z – 1 = 0, z = 1 (real)

cube roots formulae

Thus, one of the cube roots of unity is real and the other two are conjugate complex numbers.

formula of cube roots
complex cube root formula

Thus, one of the complex cube roots is the square of the other. Hence, the three cube roots of unity are 1, ω and ω2.

Complex numbers cube roots

Properties of Cube Roots of Unity:

  • Sum of the cube roots of unity is 0 that is 1 + ω + ω2 = 0.
  • Product of the cube roots of unity is 1 that is 1 . ω . ω2 = ω3 = 1.
  • Each complex cube root of unity is square of the other.
  • Each complex cube root of unity is reciprocal of other i.e. ω = 1/ω2 or ω2 = 1/ω.
Example- If α, β are the complex cube roots of unity then prove that α4 + β4 + 1/αβ = 0.

Solution- The complex cube roots of unity are ω and ω2.
Here, the complex cube roots are α and β.
∴ α = ω, β = ω2

Now, α4 + β4 + 1/αβ = ω4 + (ω2)4 + 1/ω.ω2
⇒ α4 + β4 + 1/αβ = (ω3) ω + (ω3)2 ω2 + 1
⇒ α4 + β4 + 1/αβ = ω + ω2 + 1
⇒ α4 + β4 + 1/αβ = 0
Example- If 1, ω and ω2 be the cube roots of unity then prove that (1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8)…………….to 2n factors = 22n.

Solution- (1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8) (1 – ω8 + ω16)…………….2n factors
= (1 – ω + ω2) (1 – ω2 + ω) (1 – ω + ω2) (1 – ω2 + ω)…………….2n factors
= [(1 + ω2) – ω] [(1 + ω) – ω2] [(1 + ω2) – ω] [(1 + ω) – ω2]…………….2n factors
= (-ω – ω) (-ω2 – ω2) (-ω – ω) (-ω2 – ω2)…………….2n factors
= (-2ω) (-2ω2) (-2ω) (-2ω2)…………….2n factors
= [(-2ω) (-2ω2)] [(-2ω) (-2ω2)]…………….n factor
= (4ω3) (4ω3)…………….to n factor
= (4) (4)…………….to n factor
= 4n
= (22)n
= 22n
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