## How to Find the Cube Roots of Unity?

### How to Find the Cube Roots of Unity?

The term unity refers to 1. A sequence of steps are to be followed to find the cube root of unity.

Step1:

Cube root of unity is equated to a variable say ‘z’.

$\sqrt[3]{1} = Z$

Step 2:

Cube and cube root of a number are inverse operations. So, if the cube root is shifted to the other side of the equation it becomes the cube of the number on the other side.

$1 = {Z^3}$

Step 3:

Shift ‘1’ also to the other side of the equation. So, the value on LHS will be zero.

${Z^3} – 1 = 0$

Step 4:

From the algebraic identity of ${a^3} – {b^3} = \left( {a{\text{ }} – {\text{ }}b} \right)\left( {{a^2} + {\text{ }}ab{\text{ }} + {\text{ }}{a^2}} \right)$factorize ${Z^3} – 1$. (1 can be represented as ${1^3}$ ).

$\left( {z – 1} \right)\left( {{z^2} + z + 1} \right) = 0$

Step 5:

Simplify the factors further to evaluate the value of ‘z’.

From the equation in step 5, either z – 1 = 0 or ${z^2} + z + 1 = 0$

If z – 1 =0, z = 1 (when -1 is shifted to the other side of the equation).

${z^2} + z + 1 = 0$ is simplified using the formula method of solving quadratic equations.

According to the formula,

$z = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$

In the above formula, the general form of quadratic equation is considered as

$a{x^2} + bx + c = 0$. Comparing the general equation and${z^2} + z + 1 = 0$, a = 1, b = 1 and c = 1.

Substituting these values in the formula,

So, the complex cube roots of unity obtained by solving ${z^2} + z + 1 = 0$are $– \frac{1}{2} – \frac{{\sqrt 3 }}{2}$and

$– \frac{1}{2} + \frac{{\sqrt 3 }}{2}$

However, $\sqrt { – 1} = i$ (Square root of the negative of unity is a complex imaginary number). Substituting in the roots obtained above, the three value of cube root of unity are:

$\sqrt[3]{1} = 1, – \frac{1}{2} + i\frac{{\sqrt 3 }}{2}, – \frac{1}{2} – i\frac{{\sqrt 3 }}{2}$

### Properties of Cube Root of Unity:

Property 1: There are three different values of cube root of unity among which one is a real root and the other two are complex cube roots of unity.

The real root is ‘1’ and the imaginary roots are $– \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$and $– \frac{1}{2} – i\frac{{\sqrt 3 }}{2}$.

Property 2: One of the imaginary cube roots of unity is the square root of the other.

Proof:

Let us consider one of the values of the cube root of unity which is complex is nature.

Let, $– \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$

Squaring this value of ‘z’, we get

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

Solving the expression on right side of the above equation using the algebraic identity${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we get

${Z^2} = {\left( { – \frac{1}{2}} \right)^2} + {\left( {i\frac{{\sqrt 3 }}{2}} \right)^2} + 2.\left( { – \frac{1}{2}} \right)\left( {i\frac{{\sqrt 3 }}{2}} \right)$

The above value is another value of the cube root of unity that is imaginary.

From the above proof, it can be inferred that one of the imaginary roots of unity is equal to the square of the other imaginary cube root of unity.

In general, if one imaginary root is ω, then the other root is${\omega ^2}$.

So the roots of unity are also represented as 1, $\omega$ and ${\omega ^2}$.

Property 3: Two imaginary cube roots of unity yields a product equal to unity.

Proof:

The two imaginary cube roots of unity are $– \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$and $– \frac{1}{2} – i\frac{{\sqrt 3 }}{2}$

The product will be equal to 1 when these roots are multiplied. It is solved using the algebraic identity (a + b)(a – b) = a2 – b2

∴ The product of two imaginary cube roots of unity is equal to 1. The product of imaginary cube root of unity is also represented as $\omega + {\omega ^2} = 1$

Property 4: When all the three cube roots of unity are added, the sum obtained is equal to zero.

Proof: The three roots of unity are $– \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$,$– \frac{1}{2} – i\frac{{\sqrt 3 }}{2}$. Sum of the roots is calculated as follows.

∴ The sum of cube roots of unity is equal to zero. The sum of the cube root of unity is also represented as $1 + \omega + {\omega ^2} = 0$.

Property 5: The cube of an imaginary cube root of unity is equal to one.

${\omega ^3} = 1$

Property 6: Any imaginary cube root of 1 is equal to the reciprocal of the other imaginary cube root.

Proof:

${\omega ^3} = {\omega ^2}.\omega = 1$

$\omega = \frac{1}{\omega }{\text{ }}and{\text{ }}{\omega ^2} = \frac{1}{\omega }$

### Cube Root of Unity Examples:

1. Evaluate ${\left( {1 + {\omega ^2}} \right)^3}$

Solution:

Sum of cube roots of unity is zero. i. e. $1 + \omega + {\omega ^2} = 0$

So, $1 + {\omega ^2} = – \omega$

${\left( {1 + {\omega ^2}} \right)^3} = {\left( { – \omega } \right)^3}$

$= – {\omega ^3}\left( {{\omega ^3} = 1} \right)$

=  – 1

1. Prove that ${\left( {1 + \omega } \right)^3} – {\left( {1 + {\omega ^2}} \right)^3} = {\text{ }}0$

Solution:

Sum of cube roots of unity is zero. i. e$1 + \omega + {\omega ^2} = 0$.

So, $1 + \omega = – {\omega ^2}{\text{ }}and{\text{ }}1 + {\omega ^2} = – \omega$

Substituting these values in the LHS of the question,

(1+ ω)3 – (1+ ω2)3 = (-ω2)3 – (-ω)3

=  (-ω6) – (-ω)3 3 = 1)

= -(ω3)2 – (-ω3

= -(1)2 – (-1)

= -1 + 1

= 0

1. Evaluate ${\left( {1 + \omega – {\omega ^2}} \right)^7}$if ω is one value of the cube root of unity.

Solution:

Sum of cube roots of unity is zero. i. e. 1 + ω + ω2 = 0.

So, 1 + ω = – ω2

(1 + ω – ω2)7 = (- ω2 – ω2)7

= (- 2ω2)7

= (2)7 (- ω2)7

= 128 (- ω14) 14 = ω12 . ω2)

= 128 (- ω12 . ω2) 12 = (ω3)4 and ω3 = 1

= – 128 ω2

### Fun Facts About Properties of Cube Root of Unity:

• The values of the cube root of -1 are $– 1, – \omega {\text{ }}and – {\omega ^2}.$

• Cube root of unity is also called the de Moivre number.

1. What is the Cube Root of Unity Meaning?

A. Cube root of any number is that number which when multiplied by itself twice gives that number. If the cube of any number ‘x’ is ‘y’, then it is expressed as x3 = y. Also it can be written that x = ∛y or x = y raised to the power ⅓. (i.e. x = y). The cube root of unity is that number which when raised to the power 3 gives the answer as 1. Cube roots of unity are found by using the concepts of factoring and solving quadratic equations. The value of cube root of unity are $1 – \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$,$– \frac{1}{2} – i\frac{{\sqrt 3 }}{2}$. Among the three roots of unity, two are imaginary or complex roots and one root is a real root. The real root is ‘1’ and the imaginary roots are also represented as ω and ω2.

2. What is the Value of ω3?

A. The value of the cube of any of the imaginary cube roots of ‘1’ is equal to ‘1’.

One of the properties of the cube root of unity that are imaginary is that one imaginary root is equal to the reciprocal of the other imaginary root.

i.e. $\omega = \frac{1}{{{\omega ^2}}}$

${\omega ^2} = \omega .{\text{ }}{\omega ^2} = \omega .\frac{1}{\omega } = 1$

Cube root of unity and its properties are used in solving a number of Mathematics problems which involve imaginary complex conjugate numbers. This property is also used in various calculations in physics where computational values include imaginary numbers.