# The Cube Roots of Unity

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$$^{2}$$.

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