The n^{th} roots of unity are in | vedclass

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

Question

(B) The equation for the $n^{th}$ roots of unity is given by $x^n = 1$. We can write $1$ in polar form as $\cos(2r\pi) + i\sin(2r\pi) = e^{i2r\pi}$ fo

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(B) The equation for the $n^{th}$ roots of unity is given by $x^n = 1$. We can write $1$ in polar form as $\cos(2r\pi) + i\sin(2r\pi) = e^{i2r\pi}$ for $r = 0, 1, 2, \dots, n-1$. Thus,the roots are $x = e^{i(2r\pi/n)}$. Substituting values of $r$,the roots are $1, e^{i(2\pi/n)}, e^{i(4\pi/n)}, \dots, e^{i(2(n-1)\pi/n)}$. These terms form a geometric progression $(G.P.)$ with the first term $a = 1$ and common ratio $r = e^{i(2\pi/n)}$.

The $n^{th}$ roots of unity are in

Similar Questions

If $\omega$ is a complex root of the equation $z^3 = 1$,then $\omega + \omega^{\left( \frac{1}{2} + \frac{3}{8} + \frac{9}{32} + \frac{27}{128} + \dots \right)}$ is equal to

If $i^2 = -1$,then $(1 + \sqrt{3} i)^{2022} - (\sqrt{3} - i)^{2022} = $

If $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1$ where $m, n \in N$,then the greatest common divisor of the least values of $m$ and $n$ is:

If $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$ where $x$ and $y$ are real,then the ordered pair $(x, y)$ is

If $\omega$ is a complex cube root of unity,then $\cos \left(\sum_{k=1}^7(k-\omega)(k-\omega^2) \frac{\pi}{175}\right) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module