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
⇒ 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
thus three cube root of unity are
These root are also called complex cube root
or
imaginary cube root of unity.
Cube root of unity power 2
(A)
After solving this we get
(B)
After solving we get
We see that each complex cube root of unity is the square of each other
NOTE
If then
And if
then
Sum of cube roots of unity is zero,
that is
1 + ω + ω² = 0 ……………………… A
We know that the cube root of unity
1, ,
putting these value in A
1 + +
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 =( ) (
)
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 root of unity power 15
we know that ω³ = 1
Cube roots of unity power 27
we know that ω³ = 1
Cube roots of unity power -1 (negative one)
we know that ω³ = 1
Cube roots of unity power -12 (negative twelve)
we know that ω³ = 1
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