If 1, omega, omega^2 are the cube roots of un | vedclass

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

Question

(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$ and $\omega^3=1$.We simplify the powers of $\omega

Vedclass · Accepted Answer

$7^4$

Vedclass · Answer

(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$ and $\omega^3=1$.We simplify the powers of $\omega$: $\omega^{10} = \omega^{3 	imes 3 + 1} = \omega$ and $\omega^{11} = \omega^{3 	imes 3 + 2} = \omega^2$.Substituting these into the expression:$(2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2 = (2-\omega)^2(2-\omega^2)^2(2-\omega)^2(2-\omega^2)^2$$= [(2-\omega)(2-\omega^2)]^4$$= [4 - 2(\omega+\omega^2) + \omega^3]^4$Since $\omega+\omega^2 = -1$ and $\omega^3 = 1$:$= [4 - 2(-1) + 1]^4$$= [4 + 2 + 1]^4 = 7^4$.

If $1, \omega, \omega^2$ are the cube roots of unity,then $(2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2=$

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $x^2+x+1=0$, then the equation whose roots are $\alpha^{19}$ and $\beta^7$ is

If $\omega$ is a cube root of unity,then the value of $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5$ is:

If $1, \omega, \omega^2$ are the cube roots of unity,then $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}=$

If $|Z|=2$, $Z_1=\frac{Z}{2} e^{i \alpha}$ and $\theta$ is the $\operatorname{amp}(Z)$, then $\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}}=$

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^{12}+\beta^{12}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module