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

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(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

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(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

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