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

(C) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.We simplify the expression: $(3 + 5\ome

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(C) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.We simplify the expression: $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2$.Using $3 + 3\omega + 3\omega^2 = 3(1 + \omega + \omega^2) = 3(0) = 0$,we can rewrite the terms:First term: $(3 + 3\omega + 3\omega^2 + 2\omega)^2 = (0 + 2\omega)^2 = 4\omega^2$.Second term: $(3 + 3\omega + 3\omega^2 + 2\omega^2)^2 = (0 + 2\omega^2)^2 = 4\omega^4$.Since $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$,the sum is $4\omega^2 + 4\omega$.Using $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.Therefore,$4\omega^2 + 4\omega = 4(\omega^2 + \omega) = 4(-1) = -4$.

If $\omega$ is a complex cube root of unity,then $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2 = $

Similar Questions

If $x = a$,$y = b\omega$,and $z = c\omega^2$,where $\omega$ is a complex cube root of unity,then $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \dots$

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} = $

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

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:

The number of distinct solutions of the equation $x^{11}-x^7+x^4-1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module