If 2 alpha = -1 - i sqrt{3} and 2 beta = -1 + | vedclass

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

Question

(D) Given $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$.Note that $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $\omeg

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$.Note that $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $\omega$ is the complex cube root of unity.Thus,$\alpha + \beta = -1$ and $\alpha \beta = 1$.We need to evaluate $5 \alpha^4 + 5 \beta^4 + \frac{7}{\alpha \beta}$.Since $\alpha^3 = 1$ and $\beta^3 = 1$,we have $\alpha^4 = \alpha$ and $\beta^4 = \beta$.So,$5 \alpha^4 + 5 \beta^4 + \frac{7}{\alpha \beta} = 5(\alpha + \beta) + \frac{7}{1}$.Substituting $\alpha + \beta = -1$,we get $5(-1) + 7 = -5 + 7 = 2$.

If $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$,then $5 \alpha^4 + 5 \beta^4 + 7 \alpha^{-1} \beta^{-1}$ is equal to

Similar Questions

${\left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right)^{1000}} = $

If $1, \omega, \omega^2$ are the cube roots of unity,then the roots of the equation $8z^3 - 12z^2 + 6z - 28 = 0$ are

If $\omega$ is a cube root of unity,then $(1 + \omega)^3 - (1 + \omega^2)^3 = $

If $\omega$ is an imaginary cube root of unity,then the value of $(1-\omega+\omega^{2}) \cdot(1-\omega^{2}+\omega^{4}) \cdot(1-\omega^{4}+\omega^{8}) \cdot \ldots$ ($2n$ factors) is

For $n \in N$,$\left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module