If 1, omega, omega^2, omega^3, dots, omega^{n | vedclass

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

Question

(C) Since $1, \omega, \omega^2, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,they are the roots of the equation $x^n - 1 = 0$.Therefore,we can

Vedclass · Accepted Answer

$n$

Vedclass · Answer

(C) Since $1, \omega, \omega^2, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,they are the roots of the equation $x^n - 1 = 0$.Therefore,we can write the polynomial as:$x^n - 1 = (x - 1)(x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$Dividing both sides by $(x - 1)$,we get:$\frac{x^n - 1}{x - 1} = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$Using the geometric series formula,the left side simplifies to:$x^{n-1} + x^{n-2} + \dots + x + 1 = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$Now,taking the limit as $x 	o 1$ or simply substituting $x = 1$ on both sides:$1^{n-1} + 1^{n-2} + \dots + 1 + 1 = (1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1})$Since there are $n$ terms on the left side,the sum is $n$.Thus,$(1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1}) = n$.

If $1, \omega, \omega^2, \omega^3, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,then $(1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1})$ equals

Similar Questions

The product of the four values of $(1+i \sqrt{3})^{3/4}$ is

If $\alpha$ and $\beta$ are the roots of the equation $x^2+2x+2=0$,then $\alpha^{15}+\beta^{15}=$

If ${\left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n}$ is an integer,then the smallest positive integer $n$ is

The value of $\sum_{k=1}^{6}\left(\sin \frac{2 k \pi}{7}-i \cos \frac{2 k \pi}{7}\right)$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module