If alpha_1, alpha_2, ldots, alpha_{23} are th | vedclass

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(D) Since $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the $23^{rd}$ roots of unity,they satisfy the equation $\alpha^{23} - 1 = 0$,which implies $\a

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(D) Since $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the $23^{rd}$ roots of unity,they satisfy the equation $\alpha^{23} - 1 = 0$,which implies $\alpha^{23} = 1$. Now,consider the sum $S = \alpha_1^{47} + \alpha_2^{47} + \ldots + \alpha_{23}^{47}$. Since $\alpha_k^{23} = 1$ for each $k = 1, 2, \ldots, 23$,we can write $\alpha_k^{47} = \alpha_k^{23 	imes 2 + 1} = (\alpha_k^{23})^2 \cdot \alpha_k = (1)^2 \cdot \alpha_k = \alpha_k$. Therefore,$S = \alpha_1 + \alpha_2 + \ldots + \alpha_{23}$. The sum of the $n^{th}$ roots of unity is $0$ for $n > 1$. Thus,$\alpha_1 + \alpha_2 + \ldots + \alpha_{23} = 0$.

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 $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the $23^{rd}$ roots of unity,then $\alpha_1^{47} + \alpha_2^{47} + \ldots + \alpha_{23}^{47} = $

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