If omega is a complex cube root of unity,then | vedclass

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

Question

(B) Given the expression: $P = (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors.Since $\omega^3 = 1$,we have $\omega^4 =

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the expression: $P = (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors.Since $\omega^3 = 1$,we have $\omega^4 = \omega$ and $\omega^8 = \omega^2$.Thus,the expression becomes: $P = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) \dots$ to $2n$ factors.This can be grouped into $n$ pairs of $(1 + \omega)(1 + \omega^2)$.We know that $1 + \omega + \omega^2 = 0$,so $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.Therefore,$(1 + \omega)(1 + \omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1$.Since there are $n$ such pairs,the product is $1^n = 1$.

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

Similar Questions

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

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

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + x + 1 = 0$, then $\alpha^{2026} + \beta^{2026} = $

If $\omega$ is the complex cube root of unity,then the value of $\left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module