The number of common roots among the 12^{text | vedclass

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

Question

(D) The $n^{	ext{th}}$ roots of unity are given by $e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, 2, \dots, n-1$. For $12^{	ext{th}}$ roots of unity,the val

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The $n^{	ext{th}}$ roots of unity are given by $e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, 2, \dots, n-1$. For $12^{	ext{th}}$ roots of unity,the values are $e^{i \frac{2k_1\pi}{12}} = e^{i \frac{k_1\pi}{6}}$ where $k_1 \in \{0, 1, \dots, 11\}$. For $30^{	ext{th}}$ roots of unity,the values are $e^{i \frac{2k_2\pi}{30}} = e^{i \frac{k_2\pi}{15}}$ where $k_2 \in \{0, 1, \dots, 29\}$. $A$ root is common if $e^{i \frac{k_1\pi}{6}} = e^{i \frac{k_2\pi}{15}}$,which implies $\frac{k_1}{6} = \frac{k_2}{15} \pmod{2}$. This simplifies to $\frac{k_1}{2} = \frac{k_2}{5} \pmod{2}$,or $5k_1 = 2k_2 \pmod{60}$. The number of common roots between $n^{	ext{th}}$ and $m^{	ext{th}}$ roots of unity is given by $\gcd(n, m)$. Here,$\gcd(12, 30) = 6$. Thus,there are $6$ common roots.

The number of common roots among the $12^{\text{th}}$ and $30^{\text{th}}$ roots of unity is

Similar Questions

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

If $\omega$ is a complex cube root of unity,then find the value of $\sum_{x=1}^{10} ((\omega x+2)(\omega^2 x+2)-3)$.

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2014}+z^{2015}+1=0$ are

$\frac{(-1 + i\sqrt{3})^{15}}{(1 - i)^{20}} + \frac{(-1 - i\sqrt{3})^{15}}{(1 + i)^{20}}$ is equal to

$(1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}})^5=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module