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

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Consider the expression $(x+1)(x+\omega

Vedclass · Accepted Answer

$x^3-1$

Vedclass · Answer

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Consider the expression $(x+1)(x+\omega)(x-\omega-1)$.Note that $-\omega-1 = \omega^2$ since $1 + \omega + \omega^2 = 0$.So,the expression becomes $(x+1)(x+\omega)(x+\omega^2)$.We know the identity $(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc$.Here,the roots are $-1, -\omega, -\omega^2$.Thus,$(x+1)(x+\omega)(x+\omega^2) = x^3 - (-1-\omega-\omega^2)x^2 + (\omega+\omega^2+\omega^3)x - (1)(\omega)(\omega^2)$.Since $1+\omega+\omega^2=0$,we have $-1-\omega-\omega^2=0$.Also,$\omega+\omega^2+\omega^3 = -1 + 1 = 0$.And $\omega^3 = 1$.Therefore,the expression simplifies to $x^3 - (0)x^2 + (0)x - 1 = x^3 - 1$.

If $\omega$ is a complex cube root of unity,then $(x+1)(x+\omega)(x-\omega-1)$ is equal to

Similar Questions

If $\alpha$ is a non-real root of the equation $x^6-1=0$,then $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1} = $

Let $p, q \in \mathbb{R}$ and $(1-\sqrt{3}i)^{200} = 2^{199}(p + iq)$,where $i = \sqrt{-1}$. Then $p + q + q^2$ and $p - q + q^2$ are roots of the equation:

Let ${z_1}$ and ${z_2}$ be $n^{th}$ roots of unity which are ends of a line segment that subtend a right angle at the origin. Then $n$ must be of the form

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8$ is equal to

If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 2)^3 + 27 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module