If alpha is a non-real root of the equation x | vedclass

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

Question

(D) The given equation is $x^6-1=0$,which implies $x^6=1$. Since $\alpha$ is a root,$\alpha^6=1$. Also,$\alpha 
eq 1$ because $\alpha$ is a non-real

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) The given equation is $x^6-1=0$,which implies $x^6=1$. Since $\alpha$ is a root,$\alpha^6=1$. Also,$\alpha 
eq 1$ because $\alpha$ is a non-real root (the roots of $x^6-1=0$ are $e^{i2k\pi/6}$ for $k=0, 1, 2, 3, 4, 5$,and only $k=0$ gives $x=1$). We can write the numerator as $\alpha^2(1+\alpha+\alpha^2+\alpha^3)$. Since $\alpha^6-1 = (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1) = 0$ and $\alpha 
eq 1$,we have $\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$. This implies $\alpha^5+\alpha^4+\alpha^3+\alpha^2 = -\alpha-1 = -(\alpha+1)$. Substituting this into the expression: $\frac{-(\alpha+1)}{\alpha+1} = -1$.

If $\alpha$ is a non-real root of the equation $x^6-1=0$,then $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1} = $

Similar Questions

If $\omega$ is a complex cube root of unity, then the value of the expression $2(1 + \frac{1}{\omega})(1 + \frac{1}{\omega^2}) + 3(2 + \frac{1}{\omega})(2 + \frac{1}{\omega^2}) + ... + (n + 1)(n + \frac{1}{\omega})(n + \frac{1}{\omega^2})$ is...

$\sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$ is equal to

If $z = \frac{-1-i \sqrt{3}}{2}$,then $\sum_{k=1}^{2022} \left(z^k + \frac{1}{z^k}\right)^2 = $

If $1, \omega, \omega^{2}$ are three cube roots of unity,then $(1-\omega+\omega^{2})(1+\omega-\omega^{2})$ is

If $i^2 = -1$,then $(1 + \sqrt{3} i)^{2022} - (\sqrt{3} - i)^{2022} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module