Nth Root of Unity - Definition, Properties and Examples

Nth roots of unity are significant in various fields of mathematics, including algebra, number theory, and complex analysis. The nth root of unity is effective because it is cyclic in nature. They provide a fundamental example of roots of unity, which are essential in understanding polynomial equations, symmetries, and cyclic groups.

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This Story also Contains

1. Nth root of unity
2. How to find nth root of unity?
3. The sum of nth roots of unity
4. Product of nth roots of unity
5. Nth Root of Unity in Complex Numbers
6. Nth Root of Unity Properties

In this article, we will cover the concept of the nth root of unity. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Nth root of unity

Mathematically, if ‘n’ is a positive integer, then ‘x’ is said to be an nth root of unity if it satisfies the equation $x^n=1$. Thus, this equation has n roots which are also termed as the nth roots of unity. The symbol used to denote the nth root is √n. It is a radical symbol used for square root with a little n to define the nth root. In the expression $\sqrt[n]{x}$, n is known as the index and the x is known as the radicand.

How to find nth root of unity?

We solve the nth root of unity the same as the cube root, only the value becomes n instead of 3.

Let z be the nth root of unity
$\begin{array}{rl}So,z^n& =1\\ z=(1{)}^{1/n}& =1+i(0)\\ & =(\cos 0+i\sin 0{)}^{1/n}\\ & =(\cos (2k\pi +0)+i\sin (2k\pi ++0){)}^{1/n},\text{ where }k=\text{ integer }\\ & =(\cos (2k\pi )+i\sin (2k\pi ){)}^{1/n},\end{array}$
Using the De-moivre theorem, it can be written as

Now for $k=0,1,2,\dots ,(n-1)$, we get $n$ different solutions, nth roots of unity are represented by $a^k$ where $k=0,1,2,\dots ,(n-1)$.
Let, $a=\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}$
Then, nth root of unity are ${\mathrm{a}}^k(\mathrm{k}=0,1,2,3,\dots \dots \dots \dots .(\mathrm{n}-1))$
i.e. $1,a,a^2,a^3,a^4\dots \dots \dots a^{n-1}$

So these roots of unity are in geometric progression with a common ratio α = e^2iπ/n

The sum of nth roots of unity

As all these numbers are roots of the polynomial equation:

$z^n-1=0$

Using the sum of roots relation, we can see that the sum of the roots is 0 (since there is no term with $z^{n-1}$ in the equation).

Proof:

$1+\omega +{\omega }^2+\cdots +{\omega }^{n-1}=\frac{1-{\omega }^n}{1-\omega }$

Since ${\omega }^n=1$ and $\omega \ne 1$, we substitute:

$1+\omega +{\omega }^2+\cdots +{\omega }^{n-1}=\frac{1-1}{1-\omega }=\frac{0}{1-\omega }=0$

Thus, the sum of all the nth roots of unity is 0.

Product of nth roots of unity

As all these numbers are roots of the polynomial equation:

$x^n-1=0$

Using the product of roots relation, we can see that the product of the roots will be 1 if $n$ is odd, and it will be -1 if $n$ is even.

Proof:

$1\cdot \omega \cdot {\omega }^2\cdots {\omega }^{n-1}={\omega }^{0+1+2+3+\cdots +(n-1)}$

Since the sum of the first $(n-1)$ natural numbers is:

$\sum _{k=0}^{n-1}k=\frac{n(n-1)}{2}$

we get:

$1\cdot \omega \cdot {\omega }^2\cdots {\omega }^{n-1}={\omega }^{\frac{n(n-1)}{2}}$

Since ${\omega }^n=e^{2\pi i}=1$, we substitute:

${\omega }^{\frac{n(n-1)}{2}}={\left(e^{2\pi i}\right)}^{\frac{(n-1)}{2}}=e^{\pi i(n-1)}$

Now, using Euler's identity $e^{\pi i}=-1$, we get:

$1\cdot \omega \cdot {\omega }^2\cdots {\omega }^{n-1}=(-1{)}^{n-1}$

Thus,
- If $n$ is odd, $(-1{)}^{n-1}=1$, so the product of the roots is 1 .
- If $n$ is even, $(-1{)}^{n-1}=-1$, so the product of $\downarrow$ roots is -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.

$x+iy=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n}$

From this, we have:

$x=\cos \frac{2k\pi }{n},y=\sin \frac{2k\pi }{n}$

Now,

$x^2+y^2={\cos }^2\frac{2k\pi }{n}+{\sin }^2\frac{2k\pi }{n}=1$

Hence, it satisfies the equation of the circle with origin $(0,0)$.
If a complex number is represented by $\omega$, then:

$\omega =e^{2\pi i/n}=\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}$

${\omega }^n={\left(e^{2\pi i/n}\right)}^n=e^{2\pi i}=1$

Therefore, $\omega$ is the nth root of unity.
Again, using De Moivre's Theorem, the complex numbers:

$1,\omega ,{\omega }^2,\dots ,{\omega }^{n-1}$

are the $n$th roots of unity. Thus, we can say that all the complex numbers

$1,\omega ,{\omega }^2,\dots ,{\omega }^{n-1}$

are the points in a plane and vertices of a regular $n$-sided polygon, inscribed in a unit circle.

Nth Root of Unity Properties

The properties of the nth root of unity are listed below:

• The n roots of the nth roots of unity are found on the perimeter of a circle with a radius of 1 and the origin as its center (0, 0).
• The sum of all the nth roots of unity is zero.
• The product of all the nth roots of unity is:$1\cdot \omega \cdot {\omega }^2\dots \dots \dots {\omega }^{n-1}={(-1)}^{n-1}$.

$\begin{array}{|ll|}\hline n& \text{ Primitive }n\text{ nth Roots of Unity }\\ 1& 1\\ 2& -1\\ 3& e^{2\pi i/3},e^{-2\pi i/3}\\ 4& i,-i\\ 5& e^{2\pi i/5},e^{4\pi i/5},e^{-4\pi i/5},e^{-2\pi i/5}\\ 6& e^{\pi i/3},e^{-\pi i/3}\\ 7& e^{2\pi i/7},e^{4\pi i/7},e^{6\pi i/7},e^{-6\pi i/7},e^{-4\pi i/7},e^{-2\pi i/7}\\ \hline\end{array}$

Summary

The nth root of unity is an important aspect of complex numbers. Due to its cyclic property, it helps the fast calculation of high-power complex numbers. The main applications of the nth root of unity are solving polynomial functions, Fourier transform, group theory, and number theory along with polygon studies and graphs.

Solved Examples Based On the Nth Root of Unity

Example 1: z is a complex number satisfying the equation$(z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=0$ if ${\alpha }_1,{\alpha }_2,{\alpha }_3\cdots ,{\alpha }_7$ one least non-negative positive arguments corresponding to solutions, such that ${\alpha }_1<{\alpha }_2<{\alpha }_3<{\alpha }_4<{\alpha }_5<{\alpha }_6<a_7$ then ${\alpha }_2+{\alpha }_6$ equals

Solution:

As we learned in

nth roots of unity -

$z=(1{)}^{\frac{1}{n}}\Rightarrow z=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n}$

Where $k=0,1,2,\dots \dots ,(n-1)$

Equation reduces to : $Z^7-1=0$

$\Rightarrow Z=(1{)}^{\frac{1}{7}}$

$\begin{array}{rl}& \Rightarrow Z=\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7}\text{ where }k=0,1,2,3,\cdots ,6\\ & {\alpha }_1=0,a_2=\frac{2\pi }{7},{\alpha }_3=\frac{4\pi }{7},{\alpha }_4=\frac{6\pi }{7},a_3=\frac{8\pi }{7},a_6=\frac{10\pi }{7}\text{ and }{a}_7=\frac{12\pi }{7}\\ & \therefore {\alpha }_2+{\alpha }_6=\frac{12\pi }{7}\end{array}$

Hence, the answer is $\frac{12\pi }{7}$.

Example 2: If ‘p’ and ‘q’ are distinct prime numbers, then the number of distinct imaginary numbers whichare $P^{\text{th }}$ as well as $q^{\text{th }}$ roots of unity are -

1) $m^m()p,q)$
2) $max{}^{\mathrm{mI}}(p,q$

3) 1

4) 0

Solution

As we have learned in

nth roots of unity -

$z=(1{)}^{\frac{1}{\pi }}\Rightarrow z=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n}$

Where $k=0,1,2,\dots \dots ,(n-1)$
$x^p-1=0,x^q-1=0$

$\begin{array}{rl}& x^p=1,e^{2\pi i}\cdot e^{4\pi i}\dots \dots p^{th},x^q=1,e^{2\pi i}\cdot e^{1\pi i}\cdots q^{th}\\ & x=(1{)}^{\frac{1}{p}},e^{\frac{2\pi i}{p}},e^{\frac{1\pi i}{p}}\cdots p^{th},x=(1{)}^{\frac{1}{q}},e^{\frac{2\pi i}{q}},e^{\frac{1\pi i}{q}}\cdots q^{th}\end{array}$
p and q are prime numbers

Hence their factors will be

$\begin{array}{rl}& p=1\times p\\ & q=1\times q\end{array}$

Therefore

$e^{\frac{2\pi i}{P}}\ne e^{\frac{2\pi i}{q}}$

$e^{\frac{4\pi i}{p}}\ne e^{\frac{4\pi i}{q}}$

.And so on.

Since $1=1$

and it is not an imaginary root.

Hence, the answer is the option (4).

Example 3: The value of $\sum _{i=1}^{10}(\sin \frac{2k\pi }{11}-i\cos \frac{2k\pi }{11})$ is:

Solution:

As we have learned

nth roots of unity

$z=(1{)}^{\frac{1}{n}}\Rightarrow z=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n}$

Where $k=0,1,2,\dots \dots ,(n-1)$

And

The sum of all n nth roots of unity $=0$

Now,

$\begin{array}{rl}& S=\sum _{k=1}^{10}(\sin \frac{2k\pi }{11}-i\cos \frac{2k\pi }{11})\\ & =\sum _{k=1}^{10}(-i)(\cos \frac{2k\pi }{11}-\frac{1}{i}\sin \frac{2k\pi }{11})=(-i)\cdot \sum _{k=1}^{10}(\cos \frac{2k\pi }{11}+i\sin \frac{2k\pi }{11})\end{array}$.............(i)

Now

As 11th the roots of unity are

$z=$ $e^{\frac{i2k\pi }{11}}$,for $k=0,1,2,3,\dots ..10$

$z=1$ (for $k=0)$ $e^{\frac{i2k\pi }{11}}$for $k=1,2,3,\dots ..10$

$Z=1($ for $k=0),(\cos \frac{2k\pi }{11}+i\sin \frac{2k\pi }{11})$ for $k=1,2,3,\dots .,10$

And we have, the sum of roots of unity $=0$

$\Rightarrow 1+\sum _{k=1}^{10}(\cos \frac{2k\pi }{11}+i\sin \frac{2k\pi }{11})=0$

$\Rightarrow \sum _{k=1}^{10}(\cos \frac{2k\pi }{11}+i\sin \frac{2k\pi }{11})=-1$

Putting this in (i), we get

$S=(-i)\cdot (-1)=i$

Hence, the answer is i.

Example 4: If $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation $x^4+x^3+x^2+x+1=0$, then ${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}$ is equal to:

1) -4

2) -1

3) 1

4) 4

Solution

Let $x^4+x^3+x^2+x-1=0$

$\begin{array}{rl}& \frac{x^5-1}{x-1}=0(x\ne 1)\\ & x^5-1=0\\ & x=(1{)}^{1/3}\\ & x=(\cos 2r\pi +i\sin 2r\pi {)}^{1/5}\\ & x=\cos \frac{2r\pi }{5}+i\sin \frac{21\pi }{5}\end{array}$

Putting $\mathrm{r}=0,1,2,3,4$ the five roots together is given by

$\begin{array}{rl}& \cos 0+i\sin 0=1,a=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\\ & \beta =\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5},\gamma =\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\end{array}$

$\begin{array}{rl}& \delta =\cos \frac{\theta \pi }{5}+i\sin \frac{8\pi }{5}\\ & \text{ then }{a}^{2021}+{\beta }^{2021}+y^{2021}+{\delta }^{2021}=\alpha +\beta +\gamma +\delta \end{array}$

$=-1$

using De-moivres's theorem

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

Hence, the answer is the option (2).

Example 5: If ${\mathrm{a}}_{\mathrm{r}}=\cos \frac{2\mathrm{r}\pi }{9}+i\sin \frac{2\mathrm{r}\pi }{9},\mathrm{r}=1,2,3,\dots ,i=\sqrt{-1}$, then the determinant $\left|\begin{array}{lll}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$ is equal to:

1) 1) $ag$

2) $a_1{a}_9-a_3{a}_7$

3) $a_5$

4) $a_2{a}_6-a_4{a}_8$

Solution

$a_r=e^{i\frac{2\pi r}{9}}={\alpha }^r$ where $\alpha =e^{i\frac{2\pi }{9}}$ $\mathrm{\&}$ ${\alpha }^9=1$
$\mathrm{\Delta }=\left|\begin{array}{ccc}\alpha & {\alpha }^2& {\alpha }^3\\ {\alpha }^4& {\alpha }^5& {\alpha }^6\\ {\alpha }^7& {\alpha }^8& {\alpha }^9\end{array}\right|={\alpha }^3\left|\begin{array}{ccc}\alpha & {\alpha }^2& {\alpha }^3\\ \alpha & {\alpha }^2& {\alpha }^3\\ {\alpha }^7& {\alpha }^8& {\alpha }^9\end{array}\right|=0$
Also$a_1{a}_9-a_3{a}_7={\alpha }^1{\alpha }^9-{\alpha }^3{\alpha }^7={\alpha }^{10}-{\alpha }^{10}=0$

Hence, the answer is the option 2.