Cube Root of Unity in Complex Numbers

Any number that gives the answer as 1 when it is raised to the power 3 or multiplied by itself thrice  is called the Cube root of Unity. The cube root of any number is that number which when raised to the power 3 gives the number whose cube root is to be determined. The cube root of a number is denoted by the symbol \[ \sqrt[3]{}\].

Cube root can also be denoted in index form as numbers raised to the power 1/3. The cube root of unity meaning is the cube root of ‘1’. There are three values of the cube root of unity. The two complex cube roots of unity and one real cube root are given below.

How to Find the Cube Roots of Unity?

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

Step 1:

The 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 = z³

Step 3:

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

z³ − 1 = 0

Step 4:

From the algebraic identity of a³−b³ = (a − b) (a²+ ab + a²), factorize z³ −1. (1 can be represented as 1³).

(z−1) (z² + z + 1)=0

Step 5:

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

From the equation in step 4, either z − 1 = 0 or z²+ z + 1= 0.

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

z² + 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 a quadratic equation is considered as

ax²+bx+c=0. Comparing the general equation and z² + z + 1= 0, a = 1, b = 1, and c = 1.

Substituting these values in the formula, 

\[Z = \frac{-1\pm\sqrt{1^{2} -4 \times 1 \times 1}}{2 \times 1} \]

\[ Z = \frac{-1\pm\sqrt{-3}}{2} \]

So, the complex cube roots of unity obtained by solving z² + 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 values of the 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 the 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 in nature.

Let \[ -\frac{1}{2} + i \frac{\sqrt{3}}{2}\]

Squaring this value of ‘z’, we get

\[ z^{2} = (-\frac{1}{2} + i \frac{\sqrt{3}}{2})^{2} \]

Solving the expression on right side of the above equation using the algebraic identity

(a+b)²= a²+ b²+ 2ab, we get

\[ Z^{2} = (-\frac{1}{2})^{2}\] \[+ (i \frac{\sqrt{3}}{2})^{2}\]  \[ + 2.(-\frac{1}{2})(i \frac{\sqrt{3}}{2}) \]

\[ Z^{2} = (-\frac{1}{2})^{2} + (i \frac{\sqrt{3}}{2})^{2} + 2. (-\frac{1}{2})(i \frac{\sqrt{3}}{2}) \]

\[Z^{2} = \frac{1}{4} - \frac{3}{4} - i \frac{\sqrt{3}}{2} \]

\[ Z^{2} = - \frac{1}{2} - i \frac{\sqrt{3}}{2} \]

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 ω².

So, the roots of unity are also represented as 1, ω, and ω².

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

\[ (-\frac{1}{2} + i \frac{\sqrt{3}}{2} )\] \[(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) \] = \[(-\frac{1}{2})^{2} - (i \frac{\sqrt{3}}{2})^{2} \]

\[ (-\frac{1}{2} + i \frac{\sqrt{3}}{2} \]) \[(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) \] = \[\frac{1}{4} - (-\frac{3}{4})\]

\[ (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \] \[(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) \] = \[\frac{1}{4} + \frac{3}{4} \]

\[ (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \] \[(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) \] = \[1 \]

∴ The product of two imaginary cube roots of unity is equal to 1. The product of the imaginary cube root of unity is also represented as ω + ω²=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.

Sum of the roots = \[ 1+ (-\frac{1}{2} + i \frac{\sqrt{3}}{2} \]) + (\[-\frac{1}{2} - i \frac{\sqrt{3}}{2} \]) \]

Sum of the roots = \[ 1 + (-\frac{1}{2} + -\frac{1}{2}) + (i \frac{\sqrt{3}}{2} - i \frac{\sqrt{3}}{2}) \]

Sum of the roots = \[1 + (-1)+(0)\]

Sum of the roots = \[0\]

∴ The sum of cube roots of unity is equal to zero. The sum of the cube root of unity is also represented as 

1  + ω + ω² = 0.

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

ω³= 1

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

Proof: 

ω³=ω². ω=1

ω=\[\frac{1}{ω}\] and ω²= \[\frac{1}{ω}\]

Cube Root of Unity Examples:

1. Evaluate (1 + ω²)³.

Solution:

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

So, 1 + ω² = −ω

(1 + ω²)³ = (−ω)³

=−ω³ (ω³ =1)

= − 1 

2. Prove that (1 + ω)³− (1 + ω²)³= 0.

Solution:

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

So, 1 + ω = − ω² and 1 + ω² = −ω

Substituting these values in the LHS of the question,

(1+ ω)³ − (1+ ω²)³ = (−ω²)³ − (−ω)³ 

                                =  (−ω⁶) − (−ω)³ (ω³ = 1)

                                = −(ω³)² − (−ω³) 

                                = −(1)² − (−1) 

                                = −1 + 1 

                                = 0

3. Evaluate ( 1 + ω − ω²)⁷ if ω is one value of the cube root of unity.

Solution: 

The 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) (ω¹⁴ = ω¹² . ω²)

                        = 128 (− ω¹² . ω²) (ω¹² = (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,−ω, and −ω².

• The cube root of unity is also called the de Mover number.

Conclusion

So, the cube roots of unity are the numbers that when multiplied three times or raised to the power of three provide the product 1. The cube roots of unity are the same as the cube roots of one. The number system, complex numbers, factorization, and other disciplines of mathematics utilize cube roots of unity.