## Complex Numbers: De Moivre's Theorem and nth Roots of Unity

Roots of unity are an important topic in mathematics. They are especially important in number theory, and DFT or discrete Fourier transform. A root of unity also called De moivre's number or De ulation of an n^{th} root of unity is depicted by the equation z^{n }= 1, where n is a positive integer. Let us further discuss the sum of nth roots of unity and the primitive n^{th} root of unity.

## De Moivre's Theorem

De moivre's Number or De moivre's Theorem states the relationship between any real number x and an integer n. The equation that related the two is denoted as:

$\cos (x)+i \sin (x)^{n}=\cos (n x)+i \sin (n x)$

Here I denote the imaginary unit i,e i^{2} =1. It is an important formula in mathematics because it connects complex numbers and trigonometry. If we expand the left-hand side and then compare it with the real and imaginary parts, we get expressions of cos(nx) and sin(nx) in terms of cosx and sinx.

It must be noted that the De moivre's Number or De moivre's Theorem is not valid for non-integer valThus in solving, we get the solution as

### Roots of Unity

A root of unity also called De moivre's number is a complex number when raised to some positive integer power n yields 1. Unless otherwise specified the roots of unity are defined or taken as a complex number. The statement holds true even for both 1 and -1.

## Pictorial Representation of nth Root of Unity

An n^{th} root of unity is defined by the equation z^{n} = 1, where n is a positive integer. It is also the n^{th} root of the unity formula. The equation can also be written as $\sqrt[n]{1}=z$. It must be noted that all the n roots of nth roots of unity are in Geometric Progression.

Pictorial Representation of the nth Root of Unity

### nth Roots of Unity Formula

As stated in the previous section, the n^{th} root of the unity formula is given as z^{n }= 1, or $\sqrt[n]{1}=z$.

### nth Root of Complex Number

De moivre's number r theorem can also be used to find the n^{th} roots of the complex number or equivalently the $\dfrac{1}{n}$ power.

If z is a complex number and it is written in polar form as: $z=r(\cos (\theta)+i \sin (\theta))$

For this the nth root is $r(\cos (\theta+2 k \pi)+i \sin (\theta+2 k \pi))$

Where k is an integer whose value will vary between 0 and n-1.

### How to Find the nth Root of Unity

The n^{th} root of unity can be found below. We know that the n^{th} root of unity is denoted as

Z^{n} = 1, where n is a positive integer and is the nth root of unity.

If we rewrite the above equation in polar form

Z^{n} = cos 0 + i sin 0

From the above, we know that the nth root of a complex number can be denoted as,

$Z^{n}=\cos (0+2 k \pi)+i(0+\sin 2 k \pi)$, where k is an integer

Taking nth root on both the sides, we will get.

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

## Root of Unity Examples

1. What will be the value of the fifth root of unity?

Solution: Let z^{5}^{ }= 1

Writing the above equation in polar form we get the below equation

$Z^{5}=\cos (0+2 k \pi)+i \sin (0+2 k \pi)=e^{i 2 k \pi}$ , where k = 0, 1, 2,…n-1

Hence, using De Moivre's theorem or formula we get

$Z^{\dfrac{1}{5}}=\cos \left(\dfrac{2 \pi \mathrm{k}}{\mathrm{n}}\right)+i \sin \left(\dfrac{2 \pi \mathrm{k}}{\mathrm{n}}\right)=e^{\dfrac{2 \pi \mathrm{k}}{\mathrm{n}} i}$ ; k is an integer and has values ranging from 0, 1, 2,…n-1

To get the roots, we will substitute the value of k in the formula and get the results

**For, k = 0**

Z = cos 0 + i sin 0 = 1

**For k = 1**

$Z=\cos \dfrac{2 \pi}{5}+i \sin \dfrac{2 \pi}{5}=e^{\dfrac{2 \pi}{5} i}$

**For k = 2**

$Z=\cos \dfrac{4 \pi}{5}+i \sin \dfrac{4 \pi}{5}=e^{\dfrac{4 \pi}{5} i}$

**For k = 3**

$\begin{align} &\mathrm{Z}=\cos \left(\dfrac{6 \pi}{5}\right)+i \sin \left(\dfrac{6 \pi}{5}\right)=e^{\dfrac{-4 \pi}{5} i} \\ &\because \dfrac{6 \pi}{5}-2 \pi=-\dfrac{4 \pi}{5} \end{align}$

**For k = 4**

$\begin{align} &z=\cos \left(\dfrac{8 \pi}{5}\right)+i \sin \left(\dfrac{8 \pi}{5}\right)=e^{\dfrac{-2 \pi}{5} i} \\ &\because \dfrac{8 \pi}{5}-2 \pi=-\dfrac{2 \pi}{5} \end{align}$

Therefore, the fourth roots of unity are $1,~e^{\dfrac{2 \pi}{5} i},~e^{\dfrac{4 \pi}{5} i},~e^{\dfrac{-4 \pi}{5} i},~e^{\dfrac{-2 \pi}{5} i}$

2. Calculate the Fourth root of unity by taking an example

Solution: Let, $Z = 1^{\dfrac{1}{4}}$

Z^{4} = 1

Writing the above equation in polar form we get the below equation

$z^{4}=\cos (0+2 k \pi)+i \sin (0+2 k \pi)=e^{i 2 k \pi}$ , where k = 0, 1, 2,…

Hence, using De Moibstitute the value of k in the formula and get the results

For, k = 0; Z = cos 0 + i ${\pi} { 2}+i \sin \dfrac{3 \pi} { 2}=-\cos \dfrac{\pi}{ 2}-i \sin \dfrac{\pi}{ 2}=-i$

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

### Primitive **n**^{th} Root of Unity

^{th}

Primitive n^{th} roots of unity are the ones whose multiplication order is n. A number r is called an n^{th} root of unity if r^{n }= 1 and a primitive n^{th} root of unity are if, in addition, n is the smallest integer of k = 1….. n for which r^{K} = 1.

Let us understand the Primitive n^{th} roots of unity using the example below.

The 5th root of unity is $e^{\dfrac{6\pi i}{5}}$. It can also be said as the 10th,15th, 20th or the 100th root of unity. It is because when $e^{\dfrac{6\pi i}{5}}$ is raised to any power of 5 its value is 1.

### Sum of n Roots of Unity

The sum of all the nth roots of unity is zero whereas the product of n^{th} roots of unity is (-1)^{n-1}.

The sum of all n roots of unity can be derived as follows:

Let

$\omega=e^{2 \pi i n}$

$\omega=e^{2 \pi i n}$, its roots of unity will be of the form (since it is the primitive n^{th} root of unity)

$\begin{align} &1+\omega+\omega^{2}+\omega^{3}+\cdots+\omega^{n-1}=\dfrac{\omega^{n}-1}{\omega-1} \\ &1+\omega+\omega^{2}+\omega^{3}+\cdots+\omega^{n-1}=\dfrac{1-1}{\omega-1} \\&1+\omega+\omega^{2}+\omega^{3}+\cdots+\omega^{n-1}=0 \end{align}$

Therefore, the sum of all n roots of unity is zero.

## Conclusion

The article narrates a thorough walkthrough of De Moivre’s theorem and the n^{th} roots of unity. It also discusses the sum of n^{th} roots of unity which is zero and the primitive n^{th} root of unity. The Primitive n^{th} roots of unity are the ones whose multiplication order is n. It must be noted that all the n roots of n^{th} roots unity are in Geometrical Progression or GP. Also, all the n roots of n^{th} roots unity lie on the circle with their centre at the origin and with a radius of magnitude 1. All these divide the circle into equal parts as can be seen from the figure above.

## FAQs on De Moivre's Theorem and nth Roots of Unity for JEE

1. How do we write the primitive 6th root of unity in complex form?

The nth root of are of the form $e^{\dfrac{2 \pi i k} { n}}$, where k is an integer. Therefore, for the 6th root of unity, the value of k will vary between 0 to 5 to calculate the roots.

$\begin{align} &e^{0 \}{ 3}}=-\dfrac{1}{2}-i \dfrac{\sqrt{3}}{2} \\ &e^{\dfrac{5 \pi i} { 3}}=\dfrac{1}{2}-i \dfrac{\sqrt{3}}{2} \end{align}$

In complex form, these can be written as $\left(\pm 1, \pm \dfrac{1}{2}, \pm i \dfrac{\sqrt{3}}{2}\right)$

2. What is the cube root of unity? Also, prove that the product of all the nth roots of unity is (-1)^{n-1}.

The cube root of a number is denoted as $x=3\sqrt{3}y$ or $x=y^{\dfrac{1}{3}}$. However, the cube root of unity is denoted as$z^{\dfrac{1}{3}=1}$. On calculating the values of the cube root of unity comes out to be ${1,-\dfrac{1}{2}+i}$

The product of n, nth roots of unity is (-1)^{n-1}.

$\begin{align} &1 . \omega \cdot \omega^{2} \cdot \omega^{3} \ldots \ldots \omega^{n-1}=\omega \dfrac{n(n-1)}{2}\\ &\left(\omega^{n}\right)^{\dfrac{(n-1)}{2}}=\left(e^{2 \pi i}\right)^{\frac{(n-1)}{2}}=(-1)^{(n-1)} \end{align}$

Here we have denoted the complex number as $\omega$.