If alpha is a root of z^2-z+1=0,then left(alp | vedclass

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(A) The roots of $z^2-z+1=0$ are $z = \frac{1 \pm i\sqrt{3}}{2} = -\omega^2, -\omega$,where $\omega$ is the complex cube root of unity.Let $\alpha = -

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The roots of $z^2-z+1=0$ are $z = \frac{1 \pm i\sqrt{3}}{2} = -\omega^2, -\omega$,where $\omega$ is the complex cube root of unity.Let $\alpha = -\omega$. Then $\alpha^2 = \omega^2$ and $\alpha^3 = -\omega^3 = -1$. Thus $\alpha^6 = 1$.We evaluate the terms $T_n = \alpha^n + \frac{1}{\alpha^n}$ for $n = 2014, 2015, 2016, 2017, 2018$:$n=2014: \alpha^{2014} = (\alpha^6)^{335} \cdot \alpha^4 = \alpha^4 = \omega^4 = \omega$. So $T_{2014} = \omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.$n=2015: \alpha^{2015} = \alpha^5 = \omega^5 = \omega^2$. So $T_{2015} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.$n=2016: \alpha^{2016} = \alpha^6 = 1$. So $T_{2016} = 1 + \frac{1}{1} = 2$.$n=2017: \alpha^{2017} = \alpha^1 = -\omega$. So $T_{2017} = -\omega - \frac{1}{\omega} = -(\omega + \omega^2) = 1$.$n=2018: \alpha^{2018} = \alpha^2 = \omega^2$. So $T_{2018} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.Substituting these into the expression: $(-1)^1 + (-1)^2 + (2)^3 + (1)^4 + (-1)^5 = -1 + 1 + 8 + 1 - 1 = 8$.

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	$4.$ $2$

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