Nth root of unity

Nth root of unity: Usually, the root of unity is a complex number, which is raised to power n (integer) and results in a value equal to 1. This root of the unit is also termed as the Moivre number. Thus, if n is a positive integer and Z is a value that is equal to nth root of unity, then;

Zⁿ = 1

Or

ⁿ√1 = Z

Hence, the above equation has the nth root of unity, equal to Z. All the n roots of the nth roots of unity are in GP (Geometric progression). If we add all the n roots of the nth root, then the resulting value is equal to zero. And if we multiply all the n roots of nth roots, unity is equal to (-1)^n-1.

How to Find nth Root of Unity?

As we have already discussed, if ‘Z’ is the nth root of unity, then it will satisfy the following:

Zⁿ = 1

Thus, the positive value of the integer is the nth root of unity.

Now, we can write the above equation in polar form:

Zⁿ = cos 0 + i sin 0

Zⁿ = cos (0+2kπ) + i (0+sin 2kπ) [where, k is an integer]

Now taking nth root on both sides, we get;

Z = (cos 2kπ + i sin 2kπ)^1/n

Using deMoivre’s theorem, we can find the nth root of unity.

Z = (cos (2kπ/n) + i sin (2kπ/n)) = e^(i2kπ/n) ; where k = 0 , 1, 2 , 3 , 4 , ……… , (n-1)

The above equation represents the nth root of unity, only if Zⁿ = 1.

Thus, each root of unity becomes:

Z = cos [(2kπ)/n] + i sin[(2kπ)/n] where 0 ≤ k ≤ n-1

Nth Root of Unity in Complex Numbers

The general form of a complex number is given by:

x+iy

Where ‘x’ is the real part and ‘iy’ is the imaginary part.

These complex numbers are usually plotted on the argand plane or the complex plane. Now comparing the roots of unity equation with the complex number form, we get;

x + iy = cos [(2kπ)/n] + i sin[(2kπ)/n]

x = cos [(2kπ)/n]

y = sin[(2kπ)/n]

So,

x² + y² = cos² [(2kπ)/n] + sin² [(2kπ)/n] = 1

Hence, it satisfies the equation of the circle with origin (0,0).

If a complex number is represented by ω , then;

ω = e^2πi/n = cos (2πi/n) + i sin (2πi/n)

ωⁿ = (e^2πi/n)ⁿ = e^2πi = 1

Therefore, ω is the nth root of unity.

Again, using deMoivre’s theorem, the complex numbers 1, ω, ω², …., ω^n-1 are nth roots of unity. Thus, we can say all the complex numbers 1, ω, ω², …., ω^n-1 are the points in a plane and vertices of a regular n-sided polygon, inscribed in a unit circle.

Properties of nth root of unity

• The n roots of nth roots unity lie on the circumference of the circle, whose radius is equal to 1 and centre is the origin (0,0).
• The three cube roots of unity are 1, -1/2+i√(3)/2, -1/2 – i√(3)/2
• If two imaginary cube roots are multiplied, then the product we get is equal to 1.
• One of the imaginary cube roots of unity is the square of another.
• The sum of all nth roots of unity is equal to zero. 1 + [(-1 + √3 i ) /2] + [(-1 – √3 i ) /2] = 0
• The nth roots of unity 1,ω,ω² ,… …,ω^n-1 are in geometric progression with a common ratio ω. Hence, 1+ ω + ω² +… + ω^n-1 = 1- ωⁿ/1- ω = 0 since ωⁿ = 1 and ω ≠ 1
• The product of all the nth roots of unity is: 1.ω.ω² … …ω^n-1 = (-1)^n-1

Video Lesson

Cube roots of infinity

Related Articles

• Complex Numbers in Maths
• Argand Plane & Polar Representation Of Complex Number
• Polar Form of Complex Numbers
• Argument of Complex Numbers

Solved Examples

Q.1: What are the cube roots of unity?

Solution: To find: 1^1/3 = ?

Let Z = 1^1/3

Z³ = 1

Z³ – 1 = 0

By the formula we know;

(a³ – b³) = (a – b) ( a² + ab + b²)

Therefore,

Now, (Z³ – 1³) = 0

or (Z – 1)(Z² + Z + 1) = 0

Therefore,

Z = 1 (one of the roots)

or

(Z² + a + 1) = 0

By using quadratic roots formula for the above equation, we get;

Z = [(-1) ± √(12-4.1.1)]/2

= [-1 ± √-3]/2

= -1/2 ± i√(3)/2

Therefore, the cubic roots of unity are 1, -1/2+i√(3)/2, -1/2 – i√(3)/2.

Also check: Cube Roots of Unity

Q.2: What is the fourth root of unity?

Solution: Let, Z = 1^1/4

Z⁴ = 1

Hence, we can write the above equation in polar form:

Z⁴ = cos (0 + 2kπ ) + i sin (0 + 2kπ ) = e^i2kπ ; k = 0, 1, 2,…

Hence,

Z = cos(2kπ/4) + i sin(2kπ/4) = e^i2kπ/4 ; k is an integer

Now let us put the values of k in the above equation to get the roots.

When k = 0; Z = cos 0 + i sin 0 = 1

When k = 1; Z = cos (π/2) + i sin (π/2) = i

When k = 2; Z = cos π + i sin π = -1

When k = 3; Z = cos 3π/2 + i sin 3π/2 = – cos π/2 – i sin π/2 = -i

Therefore, the fourth roots of unity are 1, i, -1 and -i.