If omega is the complex cube root of unity,th | vedclass

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

Question

(D) We know that $1+\omega+\omega^2 = 0$,which implies $1+\omega^2 = -\omega$ and $\omega^3 = 1$.Given expression: $E = (3+5\omega+3\omega^2)^2 + (3+3

Vedclass · Accepted Answer

-$4$

Vedclass · Answer

(D) We know that $1+\omega+\omega^2 = 0$,which implies $1+\omega^2 = -\omega$ and $\omega^3 = 1$.Given expression: $E = (3+5\omega+3\omega^2)^2 + (3+3\omega+5\omega^2)^2$Rearranging terms using $3(1+\omega+\omega^2) = 0$:$E = (3(1+\omega+\omega^2) + 2\omega)^2 + (3(1+\omega+\omega^2) + 2\omega^2)^2$$E = (0 + 2\omega)^2 + (0 + 2\omega^2)^2$$E = 4\omega^2 + 4\omega^4$Since $\omega^3 = 1$,we have $\omega^4 = \omega$.$E = 4\omega^2 + 4\omega = 4(\omega^2 + \omega)$Since $1+\omega+\omega^2 = 0$,we have $\omega^2 + \omega = -1$.$E = 4(-1) = -4$.

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:

Similar Questions

Let $z = \frac{1 - i \sqrt{3}}{2}$,where $i = \sqrt{-1}$. Then the value of $21 + \left(z + \frac{1}{z}\right)^{3} + \left(z^{2} + \frac{1}{z^{2}}\right)^{3} + \left(z^{3} + \frac{1}{z^{3}}\right)^{3} + \dots + \left(z^{21} + \frac{1}{z^{21}}\right)^{3}$ is .... .

For $n \in N$,if $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right)$,then $(A_1 A_2 A_3 A_4)^4 =$

If $n$ is an integer which leaves remainder $1$ when divided by $3$,then $(1+\sqrt{3}i)^n + (1-\sqrt{3}i)^n$ equals

If $(\sqrt{3}-i)^{n}=2^{n}, n \in N$,then the least possible value of $n$ is

If $z = \sin \theta - i \cos \theta,$ then for any integer $n$

Vedclass Test Series

Exam Paper Generator

Online Exam Module