Cube Root Of Unity Power

Cube root of unity power 1, 2, 3, 4,5, 6, 7, 8, 9………………..n first we find and proof root power three of unity power 1, 2, 3, and we apply that resultant value to n power of unity.

Suppose that x be a cube root of unity.

Therefore

$Cube root of unity power$

⇒ x³ = 1

⇒ x³- 1 = 0

⇒ (x – 1)(x² + x – 1) = 0

(i) x – 1 = 0

x = 1

(ii) x² + x – 1 = 0

Applying quadratic formula

$Cube root of unity power$

thus three cube root of unity are

$Cube root of unity power$

These root are also called complex cube root

imaginary cube root of unity.

Cube root of unity power 2

(A) $Cube root of unity power$

After solving this we get

$\small \left ( \frac{-1+\sqrt{3\iota }}{2} \right )^{2}$ $\small =\frac{-1-\sqrt{3\iota }}{2}$

(B) $Cube root of unity power$

After solving we get

$Cube root of unity power$ $\small =\frac{-1+\sqrt{3\iota }}{2}$

We see that each complex cube root of unity is the square of each other

NOTE

If $\small \omega =\, \frac{-1+\sqrt{3\iota }}{2}$ then $\small \small \omega^{2} =\, \frac{-1-\sqrt{3\iota }}{2}$

And if

$\small \small \small \omega =\, \frac{-1-\sqrt{3\iota }}{2}$ then $\small \small \small \omega^{2} =\, \frac{-1+\sqrt{3\iota }}{2}$

Sum of cube roots of unity is zero,

that is

1 + ω + ω² = 0 ……………………… A

We know that the cube root of unity

1, $\small \small \small \omega =\, \frac{-1-\sqrt{3\iota }}{2}$ , $\small \small \small \omega^{2} =\, \frac{-1+\sqrt{3\iota }}{2}$

putting these value in A

1 + $\small \frac{-1+\sqrt{3\iota }}{2}$ + $\small \frac{-1-\sqrt{3\iota }}{2}$

solving this we get

0/2 = 0

Hence, cube root of unity = 1 + ω + ω ² =0

cube root of unity power 3 that is ω³ = 1

That is

ω³ = 1× ω × ω² =( $\small \frac{-1+\sqrt{3\iota }}{2}$ ) ( $\small \frac{-1-\sqrt{3\iota }}{2}$ )

ω³ = 1× ω × ω² = (-1)² – (√ 3 ι) ² / 4

= 1 – (-3) /4

ω³ = 4 / 4 =1

Therefore, the product of the complex cube roots of unity ω³ = 1

Cube roots of unity power 4

we know that ω³ = 1

$Cube roots of unity power 4$

Cube root of unity power 15

we know that ω³ = 1

$cube root of unity power 15$

Cube roots of unity power 27

we know that ω³ = 1

$\small \small \omega ^{27}=\left ( \omega ^{3} \right )^{9}=\left ( 1 \right )^{9}=1$

Cube roots of unity power -1 (negative one)

we know that ω³ = 1

$\small \omega ^{-1}=\omega ^{-3}\times \omega ^{2}=\left ( \omega ^{3} \right )^{-1}\times \omega ^{2}=\omega ^{2}$

Cube roots of unity power -12 (negative twelve)

we know that ω³ = 1

$\small \omega ^{-12}=\left ( \omega ^{3} \right )^{-4}=\left ( 1 \right )^{-4}=1$

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