The product of any two of the tenth roots of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The $n$-th roots of unity are given by $z_k = e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, \dots, n-1$.For $n=10$,the roots are $\omega^k = e^{i \frac{2k

Vedclass · Accepted Answer

Always one of the tenth roots of unity

Vedclass · Answer

(A) The $n$-th roots of unity are given by $z_k = e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, \dots, n-1$.For $n=10$,the roots are $\omega^k = e^{i \frac{2k\pi}{10}}$ where $k \in \{0, 1, \dots, 9\}$.Let $z_1 = \omega^r$ and $z_2 = \omega^s$ be two such roots,where $r, s \in \{0, 1, \dots, 9\}$.The product is $z_1 z_2 = \omega^r \cdot \omega^s = \omega^{r+s} = e^{i \frac{2(r+s)\pi}{10}}$.Since $\omega^{10} = 1$,we have $\omega^{r+s} = \omega^{(r+s) \pmod{10}}$.Since $(r+s) \pmod{10}$ is always an integer in the set $\{0, 1, \dots, 9\}$,the product $z_1 z_2$ is always one of the tenth roots of unity.

The product of any two of the tenth roots of unity is

Similar Questions

${\left( \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } \right)^n} = $

If $1, \omega$ and $\omega^2$ are the cube roots of unity,then $(a+b+c)(a+b \omega+c \omega^2)(a+b \omega^2+c \omega) = $

For $n \in N$,if $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right)$,then $(A_1 A_2 A_3 A_4)^4 =$

Which of the following is a fourth root of $\frac{1}{2} + \frac{i\sqrt{3}}{2}$?

Let $\omega \neq 1$ be a cube root of unity. Then the minimum value of the set $\{|a + b\omega + c\omega^2|^2 : a, b, c \text{ are distinct non-zero integers}\}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module