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 $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\omega$:$

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\omega$:$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$Substitute these into the expression:$\sin \left\{(\omega + \omega^2) \pi - \frac{\pi}{4}ight\}$Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.Thus,the expression becomes:$\sin \left\{(-1) \pi - \frac{\pi}{4}ight\} = \sin \left(-\pi - \frac{\pi}{4}ight) = \sin \left(-\left(\pi + \frac{\pi}{4}ight)ight)$Using the property $\sin(-	heta) = -\sin(	heta)$:$-\sin \left(\pi + \frac{\pi}{4}ight) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$Therefore,the correct option is $A$.

If $\omega$ is a complex cube root of unity,then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}=$

Similar Questions

On any given arc of positive length on the unit circle $|z|=1$ in the complex plane,

One of the roots of the equation $x^{14}+x^9-x^5-1=0$ is

If $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the $23^{rd}$ roots of unity,then $\alpha_1^{47} + \alpha_2^{47} + \ldots + \alpha_{23}^{47} = $

$(1-i \sqrt{3})^{2025}=$

If $1, \omega, \omega^2$ are the cube roots of unity,then the value of $(x+y)^2+(x \omega+y \omega^2)^2+(x \omega^2+y \omega)^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module