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$^{2}$.

11 and 12 Grade Math 

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