If alpha and beta are roots of x^{2}-x+1=0, t | vedclass

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Question

(B) The given quadratic equation is $x^{2}-x+1=0$.The roots of this equation are given by $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$

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-$2$

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(B) The given quadratic equation is $x^{2}-x+1=0$.The roots of this equation are given by $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$.These roots are $-\omega$ and $-\omega^{2}$, where $\omega$ is the complex cube root of unity.Let $\alpha = -\omega$ and $\beta = -\omega^{2}$.Then, $\alpha^{2013} + \beta^{2013} = (-\omega)^{2013} + (-\omega^{2})^{2013}$.Since $2013$ is an odd number, $(-\omega)^{2013} = -\omega^{2013} = -(\omega^{3})^{671} = -(1)^{671} = -1$.Similarly, $(-\omega^{2})^{2013} = -(\omega^{2})^{2013} = -\omega^{4026} = -(\omega^{3})^{1342} = -(1)^{1342} = -1$.Therefore, $\alpha^{2013} + \beta^{2013} = -1 + (-1) = -2$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

If $\alpha$ and $\beta$ are roots of $x^{2}-x+1=0,$ then the value of $\alpha^{2013}+\beta^{2013}$ is

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