Mathematics : Complex Numbers: The nth roots of unity

The *n*th roots of unity

The solutions of the equation *z**n** *= 1 , for positive values of integer *n *, are the *n *roots of the unity.

In polar form the equation *z**n** *= 1 can be written as

zn = cos (0 + 2kÏ€) + *i *sin (0 + 2kÏ€) = ei2kÏ€ , k = 0, 1, 2,â€¦..

Using deMoivreâ€™s theorem, we find the *n*th roots of unity from the equation given below:

Given a positive integer *n *, a complex number *z *is called an *n** *th root of unity if and only if *z**n** *= 1.

If we denote the complex number by *Ï‰ *, then

Therefore Ï‰ is an nth root of unity. From equation (1), the complex numbers 1,Ï‰,Ï‰2 ,... ...,Ï‰n-1 are nth roots of unity. The complex numbers 1,Ï‰,Ï‰2 ,... ...,Ï‰n-1 are the points in the complex plane and are the vertices of a regular polygon of n sides inscribed in a unit circle as shown in Fig 2.45. Note that because the n th roots all have the same modulus 1, they will lie on a circle of radius 1 with centre at the origin. Furthermore, the n roots are equally spaced along the circle, because successive n th roots have arguments that differ by 2Ï€/n .

The *n*th roots of unity 1,Ï‰,Ï‰2 ,... ...,Ï‰n-1 are in geometric progression with common ratio Ï‰

Therefore 1+ Ï‰ + Ï‰2 +â€¦ + Ï‰n-1 = 1- Ï‰n / 1- Ï‰ = 0 since Ï‰n = 1 and Ï‰ â‰ 1 .

The sum of all the nth roots of unity is

1+ Ï‰ + Ï‰2 +â€¦ + Ï‰n-1 = 0

The product of *n*, *n*th roots of unit is

1Ï‰Ï‰2 ... ...Ï‰n-1 = Ï‰0+1+2+3+... ...+(n-1) = Ï‰[(n-1)n]/2

The product of all the *n*th roots of unity is

1Ï‰Ï‰2 ... ...Ï‰n-1 = (-1)n-1

Note

(1) All the n roots of *n*th roots unity are in Geometrical Progression

(2) Sum of the n roots of *n*th roots unity is always equal to zero.

(3) Product of the n roots of *n*th roots unity is equal to (-1)n-1 .

(4) All the n roots of *n*th roots unity lie on the circumference of a circle whose centre is at the origin and radius equal to 1 and these roots divide the circle into n equal parts and form a polygon of n sides.

Example 2.32

Find the cube roots of unity.

Solution

We have to find 11/3 . Let z = 11/3 then z3 = 1.

In polar form, the equation z3 = 1 can be written as

z3 = cos(0 + 2kÏ€) + *i* sin(0 + 2*k*Ï€) = ei2kÏ€ , k = 0, 1, 2,...

Example 2.33

Find the fourth roots of unity.

Solution

We have to find 11/4. Let z =11/4 . Then z4 = 1 .

In polar form, the equation z4 = 1 can be written as

z4 = cos (0 + 2kÏ€ ) + *i* sin (0 + 2kÏ€ ) = ei2kÏ€ , *k* = 0, 1, 2,...

Note

(i) In this chapter the letter Ï‰ is used for nth roots of unity. Therefore the value of Ï‰ is depending on n as shown in following table.

(ii) The complex number zeiÎ¸ is a rotation of z by Î¸ radians in the counter clockwise direction about the origin.

Example 2.34

Solve the equation z3 + 8*i* = 0 , where z âˆˆ C.

Let z3 + 8*i* = 0 . Then, we get

z3 = -8*i*

Find all cube roots of âˆš3 + *i *

Solution

We have to find (âˆš3 + *i*)1/3. Let z3 = âˆš3 + *i* = r (cosÎ¸ + *i* sinÎ¸ )

Suppose z1 , z2, and z3 are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z1 = 1+ iâˆš3 , then find z2 and z3.

|z| = 2 represents the circle with centre (0, 0) and radius 2.

Let A, B, and C be the vertices of the given triangle. Since the vertices z1 , z2 , and z3 form an equilateral triangle inscribed in the circle |z| = 2 , the sides of this triangle AB, BC, and CA subtend 2Ï€/3 radians (120 degree) at the origin (circumcenter of the triangle).

(The complex number z eiÎ¸ is a rotation of z by Î¸ radians in the counter clockwise direction about the origin.)

Therefore, we can obtain z2 and z3 by the rotation of z1 by 2Ï€/3 and 4 Ï€/3 respectively.

Given that

Therefore, z2 = -2, and z3 = 1- *i*âˆš3.

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12th Mathematics : UNIT 2 : Complex Numbers : The nth roots of unity | Definition, Formula, Solved Example Problems

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