MEANING & PROPERTIES OF CUBE ROOTS OF UNITY – MINDSPARK

Many of you might have the opinion that only ‘1’ is the cube root of ‘1’ but no, that’s not the case. There are other cube roots of ‘1’ also. In this article, we are going to learn about the meaning, properties and some illustrations of the cube roots of unity. 

CUBE ROOTS OF UNITY

MEANING 

When a number is raised to the power of 3 and it results in getting the number ‘1’, then it is said to be a cube of unity and the inverse of it is the cube root of unity. In simple words, we need to determine those numbers whose cube will give the result ‘1’. The cube roots of unity are 1, $\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}$. Here ‘1’ is the real root and

$\frac{-1+\sqrt{3 i}}{2} \& \frac{-1-\sqrt{3 i}}{2}$ are imaginary roots.

HOW DO YOU FIND THE ROOTS OF UNITY?

Suppose ‘t’ is the number whose cube is 1.

$\text { So, } t^{3}=1, \quad t^{3}-1^{3}=0$

$(t-1)\left(t^{2}+t+1\right)=0 \quad\left[a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$t-1=0, \quad t^{2}+t+1=0$

 So, t = 1

For the equation $t^{2}+t+1=0$, we will use the quadratic roots formula.

We know that the formula to determine the root is    $x=\frac{-b \pm \sqrt{d}}{2 a}$,   where $d=b^{2}-4 a c$.

The values of a, b and c here are 1, 1 and 1 respectively. So, $d=1^{2}-4=-3$.

Therefore, $t=\frac{-1 \pm \sqrt{-3}}{2}$

We know that the value √-1 = iota($i$)

Hence, the two imaginary roots of unity are $\frac{-1+\sqrt{3} i}{2} \& \frac{-1-\sqrt{3} i}{2} \text { where } \omega=\frac{-1+\sqrt{3} i}{2} \text { and } \omega^{2}=\frac{-1-\sqrt{3} i}{2} .$

PROPERTIES OF THE CUBE ROOT OF UNITY

Property 1:  There are three cube roots of unity of which one is a real number and the other two are conjugate complex or imaginary numbers.

The three cube roots are 1 which is a real number and $\frac{-1+\sqrt{3} i}{2} \& \frac{-1-\sqrt{3} i}{2}$are conjugate complex or imaginary numbers.

Property 2: Square of one complex number cube root of unity is equal to the other complex number cube root of unity.

 It can be proved with the help of cube roots. We know that the cube roots of unity are 1,

$\left(\frac{-1+\sqrt{3} i}{2}\right)^{2}=\frac{(-1)^{2}+(\sqrt{3} i)^{2}-2 \sqrt{3} i}{2^{2}}$
$=\frac{1+3 i^{2}-2 \sqrt{3} i}{4}$
$=\frac{1-3-2 \sqrt{3} i}{4}$
$=\frac{2(-1-\sqrt{3} i)}{4}$
$=\frac{-1-\sqrt{3} i}{2}$

It means that ω $=\frac{-1+\sqrt{3 i}}{2}$and ${\omega}^2=\frac{-1-\sqrt{3 i}}{2}$, 1, ω and ${\omega}^2$ are the cube roots of unity.

Hence, it is proved that the square of one complex cube root of unity is equal to the other complex cube root of unity. 

Property 3:The product of the two imaginary cube roots is 1 or, the product of three cube roots of unity is 1.

We can verify it, the three cube roots of unity are 1, $\frac{-1+\sqrt{3 i}}{2}, \frac{-1-\sqrt{3 i}}{2} .$

So, $1 \times \frac{-1+\sqrt{3 i}}{2} \times \frac{-(1+\sqrt{3 i})}{2}$

= $\frac{4}{4}= 1$

We know that the product of cube roots is equal to 1. 

Therefore, $\omega^{3}=1$ & the value of ⍵ is 1.

Property 4: The addition of all the cube roots of unity is zero i.e.,$1+\omega+\omega^{2}=0$

We know that, the sum of the three cube roots of unity $=1+\omega+\omega^{2}$ 

So, $1+\frac{-1+\sqrt{3 i}}{2}+\frac{-1-\sqrt{3 i}}{2}$

Taking LCM,

We will get $\frac{2-1+\sqrt{3 i}-1-\sqrt{3 i}}{2}$, which is equal to 0.

ILLUSTRATION

Q1: Find the value of the following :$\left(1+\omega-\omega^{2}\right)\left(1-\omega+\omega^{2}\right)$  

Sol. We need to do the multiplication to find its value,

$=\left(1+\omega-\omega^{2}\right)\left(1-\omega+\omega^{2}\right)$
$=1-\omega+\omega^{2}+\omega-\omega^{2}+\omega^{3}-\omega^{2}+\omega^{3}-\omega^{4}$
$=1-\omega^{2}-\omega^{4}+2 \omega^{3}$
$=1+2-\omega^{2}\left(1+\omega^{2}\right) \quad\left[\omega^{3}=1\right]$
$=3-\omega^{2}(-\omega) \quad\left[1+\omega+\omega^{2}=0,1+\omega^{2}=-\omega\right]$
$=3+\omega^{3}$
$=4$

Q2. If $(1+\omega)^{7}=C+D \omega$, then find the value of C and D using the cube root of unity. (⍵ = 1)

Sol.

Given: $(1+\omega)^{7}=C+D \omega$

$(1+\omega)^{6}(1+\omega)=C+D \omega$

${\left[(1+\omega)^{(2)^{3}}\right](1+\omega)=C+D \omega}$

$(1+\omega^2 +2\omega)^{3}(1+\omega )=C+D\omega$

$(-\omega+2 \omega)^{3}(1+\omega)=C+D \omega$

$\omega^{3}(1+\omega)=C+D \omega    [\text{Since }\omega^{3}=1]$

By comparing LHS and RHS,

The value of C and D is 1 each.