n^th Roots of Unity

 

In general, the term root of unity which is also termed as de Moivre number is basically a complex number which, when raised to some integer n gives the result as 1.

Mathematically, if n is a positive integer, then ‘x’ is said to be an n^th root of unity if it satisfies the equation xⁿ = 1. Thus, this equation has n roots which are also termed as the nth roots of unity.
 

How do We Find the n^th Root of Unity?

As stated above, if x is an nth root of unity, then it satisfies the relation xⁿ = 1.

Now, ‘1’ can also be written as cos 0 + i sin 0.   

So, we have, xⁿ = 1

                       = cos 0 + i sin 0

                       = cos 2kπ + i sin 2kπ   [k is an integer]

Taking the nth root on both sides, we get

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

So each root of unity is

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

 The complex numbers are in the form x+iy and are plotted on the argand or the complex plane. Also, since the roots of unity are in the form cos [(2kπ)/n] + i sin[(2kπ)/n], so comparing it with the general form of complex number, we obtain the real and imaginary parts as

x = cos[(2kπ)/n] ,  y = sin[(2kπ)/n].
Now, we see that the values of x and y satisfy the equation of unit circle with centre (0,0) i.e. x² + y² = 1.

Now, we consider some of the key results on n^th roots of unity-

• The cube roots of unity are 1, (-1+i√3)/2, (-1-i√3)/2.
• If w is one of the imaginary cube root of unity then 1+ w + w² = 0. In general 1 + wⁿ + w²ⁿ = 0; where n ∈ I but is not the multiple of 3.
• In polar form the cube roots of unity are:

cos 0 + i sin 0; cos 2π/3 + isin 2π/3; cos 4π/3 + I sin 4π/3.

• Product of the n^th roots of any complex number z is z(-1)^n-1

The three cube roots of unity when plotted on the argand plane constitute the vertices of an equilateral triangle.

  

• The following factorization should be remembered:

        (a, b, c ∈ R and w is the cube root of unity)

        a³ - b³= (a-b) (a-wb) (a-x²b);

        x² +x+1=(x-w)(x-w²);

        a³ + b³= (a+b) (a+wb) (a+x²b);

        a² +ab+b² = (a-bw)(a-bw²)

        a³ + b³ + c³ - 3abc = (a+b+c) (a+wb+w²c) (a+w²b+wc);

View the video for more on n^th roots of unity

n^th Roots of unity -

If 1, α₁,  α₂,......α_n-1 are the n, n^th roots of unity then:

• They are in G.P. with common ratio i(2π/n)
• 1^p + α₁^p + α₂^p + .... α_n-1^p = 0 if p is not an integral multiple of n.
• (1 - α₁)(1 - α₂).......(1 - α_n-1) = n and,
• (1 + α₁)(1 + α₂).......(1 + α_n-1) = 0 if n is even and

                                               = 1 if n is odd.

• α₁ . α_{2 .} α₃ ...... α_n-1 = 1 or -1 according as n is odd or even.

The sum of the following series should be remembered:

(i) cos θ + cos 2θ + cos 3θ + ..................... + cos nθ

          = (sin (n^θ/2)/sinθ/2) cos((n+1)/2)^θ

(ii) sin θ + sin 2θ + ..................... + sin nθ

         = (sin(n^θ/2)/sinθ/2) sin((n+1)/2)^θ

Remark: If θ = 2π/2 then the sum of the above series vanishes.

Logarithm of a complex quantity -

(i) Log_e(α + iβ) = 1/2 log_e(α² + β²) + i(2nπ + tan^-1(β/α))  where n ∈ I.

(ii) iⁱ represents a set of positive real numbers given by e^{-(2nπ + π/2)}, n ∈ I.

Illustration: Find the smallest positive integer n for which [(1+i)/ (1-i)]ⁿ  = 1. (1980)

Solution: The given expression is [(1+i)/ (1-i)]ⁿ = 1

First, we rationalize the given expression by multiplying and dividing the given expression by the conjugate of denominator,

[(1+i)/(1-i) . (1+i)/(1+i)]ⁿ = 1

This means (2i/2)ⁿ = 1

This gives iⁿ =1.

The smallest positive integer n for which iⁿ =1 is 4.

Hence, the value of n is 4.

Illustration: Let z₁ and z₂ be nth roots of unity which subtend a right angle at the origin, then what should be the form of n if k is an integer?

Solution: We know that by the formula of argument,

arg z₁/ z₂ = π/2

Hence, z₁/ z₂ = cos π/2 + isin π/2 = i

So, z₁ⁿ/ z₂ⁿ = (i)ⁿ = 1

Hence, n = 4k.

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