If alpha is a complex number satisfying the e | vedclass

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

Question

(A) The given equation is $\alpha^{2}+\alpha+1=0$. This is the characteristic equation for the cube roots of unity,where $\alpha$ is either $\omega$ o

Vedclass · Accepted Answer

$\alpha$

Vedclass · Answer

(A) The given equation is $\alpha^{2}+\alpha+1=0$. This is the characteristic equation for the cube roots of unity,where $\alpha$ is either $\omega$ or $\omega^{2}$. We know that $\omega^{3}=1$. For $\alpha=\omega$,we have $\alpha^{31} = \omega^{31} = (\omega^{3})^{10} \cdot \omega = 1^{10} \cdot \omega = \omega = \alpha$. For $\alpha=\omega^{2}$,we have $\alpha^{31} = (\omega^{2})^{31} = \omega^{62} = (\omega^{3})^{20} \cdot \omega^{2} = 1^{20} \cdot \omega^{2} = \omega^{2} = \alpha$. In both cases,$\alpha^{31} = \alpha$.

If $\alpha$ is a complex number satisfying the equation $\alpha^{2}+\alpha+1=0$,then $\alpha^{31}$ is equal to

Similar Questions

What is the product of all the values of $(1-i)^{\frac{2}{5}}$?

What is the value of $(1-i \sqrt{3})^9$?

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

If $Z = r(\cos \theta + i \sin \theta), (\theta \neq -\pi / 2)$ is a solution of $x^3 = i$,then $r^9(\cos \theta + i \sin \theta)^9 = $

If $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right)$ for $r = 1, 2, 3, \ldots$,then the product $Z_1 Z_2 Z_3 \ldots \infty$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module