n, nth roots of unity form a series in

Concept:  

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are equal.

For example, if the first term is a and the difference between consecutive terms is d, then the arithmetic progression consisting of n terms is given by,

{a, a + d, a + 2d, a + 3d, …, a + (n - 1)d}

A geometric progression (GP) is a sequence where the ratios between every two consecutive terms are equal.

For example, if the first term is a and the ratio between consecutive terms is r, then the geometric progression consisting of n terms is given by,

{a, ar, ar², ar³, …, ar^n-1}

A harmonic progression (HP) is a sequence where the reciprocals of subsequent terms are part of an arithmetic progression.

For example, if the first term is a, b, c, d,… are considered to be the terms of an arithmetic progression, then 1/a, 1/b, 1/c, 1/d,… are the terms of a harmonic progression.

An arithmetic-geometric progression is a sequence where each term is obtained by multiplying a GP with the terms of an AP.

For example, if the first term is a, the difference is d and the ratio is r, then the arithmetic-geometric progression consisting of n terms is given by,

{a, (a + d)r, (a + 2d)r², (a + 3d)r³, …, (a + (n - 1)d)r^n-1}

As per Euler’s Formula, we can write, e^ix = cos x + i sin x

As per De Moivre’s Theorem, for any integer n, (cos x + i sin x)ⁿ = cos nx + i sin nx  

Calculation:

For any positive integer n, a complex number z is called the nth root of unity if zⁿ = 1.

We know that 1 can be written as eⁱ⁰.

Hence the nth root of one can be written as (1)^1/n = (eⁱ⁰)^1/n

We know that from Euler’s formula, the nth root of unity can be written as,

(1)^1/n = (eⁱ⁰)^1/n = (cos0 + isin0)^1/n = (cos(0 + 2kπ) + i sin(0 + 2kπ))^1/n = (cos 2kπ + isin 2kπ)^1/n  ----- (1)

Where k is any integer.

From De Moivre’s Theorem, (1) can be rewritten as below:

(cos 2kπ + i sin 2kπ)^1/n = cos \(\frac{2k\pi}{n}\) + i sin \(\frac{2k\pi}{n}\)

Hence, the nth root of unity (1)^1/n = cos \(\frac{2k\pi}{n}\) + i  sin \(\frac{2k\pi}{n}\),

The value of k will vary in the range of 0 to n - 1 for a fixed value of n.

Hence, different values of the roots of unity for a fixed value of n will be written as,

1, cos \(\frac{2\pi}{n}\) + i sin \(\frac{2\pi}{n}\), cos \(\frac{4\pi}{n}\) + i sin \(\frac{4\pi}{n}\), cos \(\frac{6\pi}{n}\) + i sin \(\frac{6\pi}{n}\), …, cos \(\frac{2(n-1)\pi}{n}\) + i sin \(\frac{2(n-1)\pi}{n}\)

If cos \(\frac{2\pi}{n}\) + i sin \(\frac{2\pi}{n}\) = x, then the different roots are, for k = 1, 2, 3, 4.....

cos \(\frac{2\pi \times 0}{n}\) + i sin \(\frac{2\pi \times 0}{n}\) =  (cos \(\frac{2\pi}{n}\) + i sin \(\frac{2\pi}{n}\))⁰ = x⁰  

cos \(\frac{2\pi}{n}\)+ i sin \(\frac{2\pi}{n}\) = x

cos \(\frac{4\pi}{n}\)+ i sin \(\frac{4\pi}{n}\) = (cos \(\frac{2\pi}{n}\) + i sin \(\frac{2\pi}{n}\))² = x²

cos \(\frac{6\pi}{n}\) + i sin \(\frac{6\pi}{n}\) = (cos \(\frac{2\pi}{n}\) + i sin \(\frac{2\pi}{n}\))³ = x³

cos \(\frac{8\pi}{n}\) + i sin \(\frac{8\pi}{n}\) = (cos \(\frac{2\pi}{n}\) + i sin \(\frac{2\pi}{n}\))⁴ = x⁴

and, so on.

We can see that each root is ‘x’ times the previous root, hence maintaining the same ratio.

Hence, the nth roots of unity form a geometric progression.

Hence, the correct answer is option 2.