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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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.
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.
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{aligned}
S o, z^n & =1 \\
z=(1)^{1 / n} & =1+i(0) \\
& =(\cos 0+i \sin 0)^{1 / n} \\
& =(\cos (2 k \pi+0)+i \sin (2 k \pi++0))^{1 / n}, \text { where } k=\text { integer } \\
& =(\cos (2 k \pi)+i \sin (2 k \pi))^{1 / n},
\end{aligned}
$
Using the De-moivre theorem, it can be written as
Now for $k=0,1,2, \ldots,(n-1)$, we get $n$ different solutions, nth roots of unity are represented by $a^k$ where $k=0,1,2, \ldots,(n-1)$.
Let, $a=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}$
Then, nth root of unity are $\mathrm{a}^k(\mathrm{k}=0,1,2,3, \ldots \ldots \ldots \ldots .(\mathrm{n}-1))$
i.e. $1, a, a^2, a^3, a^4 \ldots \ldots \ldots a^{n-1}$
So these roots of unity are in geometric progression with a common ratio α = e2iπ/n
As all these numbers are roots of the polynomial equation
$z^n-1=0$
So, using the sum of roots relation, we can see that the sum of roots = 0 (As there is no term with zn-1 in the equation)
As all these numbers are roots of the polynomial equation
$2^n-1=0$
So, using the product of roots relation, we can see that the product of roots will be 1 if n is odd and it will be (-1) if n is even.
The properties of the nth root of unity are listed below:
Solved Examples Based On the Nth Root of Unity:
Example 1: z is a complex number satisfying the equation$(z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0$ 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 -
$z=(1)^{\frac{1}{n}} \Rightarrow z=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}$
Where $k=0,1,2, \ldots \ldots,(n-1)$
Equation reduces to : $Z^7-1=0$
$\Rightarrow Z=(1)^{\frac{1}{7}}$
$\begin{aligned} & \Rightarrow Z=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7} \text { where } k=0,1,2,3, \cdots, 6 \\ & \alpha_1=0, a_2=\frac{2 \pi}{7}, \alpha_3=\frac{4 \pi}{7}, \alpha_4=\frac{6 \pi}{7}, a_3=\frac{8 \pi}{7}, a_6=\frac{10 \pi}{7} \text { and } a_7=\frac{12 \pi}{7} \\ & \therefore \alpha_2+\alpha_6=\frac{12 \pi}{7}\end{aligned}$
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) $\left.m^m() p, q\right)$
2) $\max { }^{\operatorname{mI}}(p, q$
3) 1
4) 0
Solution
As we have learned in
nth roots of unity -
$
z=(1)^{\frac{1}{\pi}} \Rightarrow z=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}
$
Where $k=0,1,2, \ldots \ldots,(n-1)$
$x^p-1=0, x^q-1=0$
$
\begin{aligned}
& x^p=1, e^{2 \pi i} \cdot e^{4 \pi i} \ldots \ldots p^{t h}, x^q=1, e^{2 \pi i} \cdot e^{1 \pi i} \cdots q^{t h} \\
& x=(1)^{\frac{1}{p}}, e^{\frac{2 \pi i}{p}}, e^{\frac{1 \pi i}{p}} \cdots p^{t h}, \quad x=(1)^{\frac{1}{q}}, e^{\frac{2 \pi i}{q}}, e^{\frac{1 \pi i}{q}} \cdots q^{t h}
\end{aligned}
$
p and q are prime numbers
Hence their factors will be
$
\begin{aligned}
& p=1 \times p \\
& q=1 \times q
\end{aligned}
$
Therefore
$e^{\frac{2 \pi i}{P}} \neq e^{\frac{2 \pi i}{q}}$
$e^{\frac{4 \pi i}{p}} \neq e^{\frac{4 \pi i}{q}}$
.
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 $\sum_{i=1}^{10}\left(\sin \frac{2 k \pi}{11}-i \cos \frac{2 k \pi}{11}\right)$ is:
Solution:
As we have learned
nth roots of unity
$z=(1)^{\frac{1}{n}} \Rightarrow z=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}$
Where $k=0,1,2, \ldots \ldots,(n-1)$
And
The sum of all n nth roots of unity $=0$
Now,
$\begin{aligned} & S=\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}-i \cos \frac{2 k \pi}{11}\right) \\ & =\sum_{k=1}^{10}(-i)\left(\cos \frac{2 k \pi}{11}-\frac{1}{i} \sin \frac{2 k \pi}{11}\right)=(-i) \cdot \sum_{k=1}^{10}\left(\cos \frac{2 k \pi}{11}+i \sin \frac{2 k \pi}{11}\right)\end{aligned}$.............(i)
Now
As 11th the roots of unity are
$z=$ $e^{\frac{i 2 k \pi}{11}}$,for $k=0,1,2,3, \ldots . .10$
$z=1$ (for $k=0)$ $e^{\frac{i 2 k \pi}{11}}$for $k=1,2,3, \ldots . .10$
$Z=1($ for $k=0),\left(\cos \frac{2 k \pi}{11}+i \sin \frac{2 k \pi}{11}\right)$ for $k=1,2,3, \ldots ., 10$
And we have, the sum of roots of unity $=0$
$\Rightarrow 1+\sum_{k=1}^{10}\left(\cos \frac{2 k \pi}{11}+i \sin \frac{2 k \pi}{11}\right)=0$
$\Rightarrow \sum_{k=1}^{10}\left(\cos \frac{2 k \pi}{11}+i \sin \frac{2 k \pi}{11}\right)=-1$
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)
2)
3)
4)
Solution
Let $x^4+x^3+x^2+x-1=0$
$\begin{aligned} & \frac{x^5-1}{x-1}=0 \quad(x \neq 1) \\ & x^5-1=0 \\ & x=(1)^{1 / 3} \\ & x=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 5} \\ & x=\cos \frac{2 r \pi}{5}+i \sin \frac{21 \pi}{5}\end{aligned}$
Putting $\mathrm{r}=0,1,2,3,4$ the five roots together is given by
$\begin{aligned} & \cos 0+i \sin 0=1, \quad a=\cos \frac{2 \pi}{5}+i \sin \frac{2 \pi}{5} \\ & \beta=\cos \frac{4 \pi}{5}+i \sin \frac{4 \pi}{5}, \gamma=\cos \frac{6 \pi}{5}+i \sin \frac{6 \pi}{5}\end{aligned}$
$\begin{aligned} & \delta=\cos \frac{\theta \pi}{5}+i \sin \frac{8 \pi}{5} \\ & \text { then } a^{2021}+\beta^{2021}+y^{2021}+\delta^{2021}=\alpha+\beta+\gamma+\delta\end{aligned}$
$=-1$
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 $\mathrm{a}_{\mathrm{r}}=\cos \frac{2 \mathrm{r} \pi}{9}+i \sin \frac{2 \mathrm{r} \pi}{9}, \mathrm{r}=1,2,3, \ldots, i=\sqrt{-1}$, 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) $a g$
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 $\alpha=e^{i \frac{2 \pi}{9}}$ $\&$ $\alpha^9=1$
$\Delta=\left|\begin{array}{ccc}\alpha & \alpha^2 & \alpha^3 \\ \alpha^4 & \alpha^5 & \alpha^6 \\ \alpha^7 & \alpha^8 & \alpha^9\end{array}\right|=\alpha^3\left|\begin{array}{ccc}\alpha & \alpha^2 & \alpha^3 \\ \alpha & \alpha^2 & \alpha^3 \\ \alpha^7 & \alpha^8 & \alpha^9\end{array}\right|=0$
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.
Complex numbers are the numbers in which complex or imaginary parts exist. It is represented as a+ib.
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 third root of unity is also known as the cube roots of unity and these are$1$ $e^{\frac{2 \pi}{3} i}$, $e^{-\frac{2 \pi}{3} i}$.
Yes, 1 is an nth root of unity.
The sum of nth root of unity is 0.