If alpha and beta are the roots of the equati | vedclass

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

Question

(C) Given the quadratic equation $x^2-x+1=0$. The roots are $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ and $-\omega^2$,whe

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given the quadratic equation $x^2-x+1=0$. The roots are $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ and $-\omega^2$,where $\omega$ is the complex cube root of unity. Let $\alpha = -\omega$ and $\beta = -\omega^2$. We need to calculate $\alpha^{2009} + \beta^{2009} = (-\omega)^{2009} + (-\omega^2)^{2009}$. This simplifies to $-(\omega^{2009} + \omega^{4018})$. Since $\omega^3 = 1$,we have $\omega^{2009} = \omega^{3 	imes 669 + 2} = \omega^2$ and $\omega^{4018} = \omega^{3 	imes 1339 + 1} = \omega$. Thus,$\alpha^{2009} + \beta^{2009} = -(\omega^2 + \omega)$. Using the identity $1 + \omega + \omega^2 = 0$,we get $\omega^2 + \omega = -1$. Therefore,$\alpha^{2009} + \beta^{2009} = -(-1) = 1$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$,then $\alpha^{2009}+\beta^{2009}$ is equal to

Similar Questions

If $\cos \theta + i \sin \theta, \theta \in R$, is a root of the equation $a_0 x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n = 0$, where $a_0, a_1, \ldots, a_n \in R$ and $a_0 \neq 0$, then the value of $a_1 \sin \theta + a_2 \sin 2 \theta + \ldots + a_n \sin n \theta$ is:

If $z = \left(\frac{\sqrt{3}+i}{2}\right)^5 + \left(\frac{\sqrt{3}-i}{2}\right)^5$, then

The smallest positive integral value of $n$ such that $\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{n} = 1$ is:

If $\alpha$ is a non-real root of $x^7=1$,then $\alpha(1+\alpha)(1+\alpha^2+\alpha^4) = $

The real part of $(\cos 4 + i \sin 4 + 1)^{2020}$ is $.........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module