Let z_k = cos left(frac{2kpi}{10}right) + i s | vedclass

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(C) $(P)$ Since $z_k = e^{i(2k\pi/10)}$,we have $z_k \cdot z_j = e^{i(2(k+j)\pi/10)} = 1$ if $k+j = 10$. For any $k \in \{1, \ldots, 9\}$,we can choos

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$1 \quad 2 \quad 3 \quad 4$

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(C) $(P)$ Since $z_k = e^{i(2k\pi/10)}$,we have $z_k \cdot z_j = e^{i(2(k+j)\pi/10)} = 1$ if $k+j = 10$. For any $k \in \{1, \ldots, 9\}$,we can choose $j = 10-k \in \{1, \ldots, 9\}$. Thus,the statement is True $(1)$.$(Q)$ The equation $z_1 \cdot z = z_k$ is a linear equation in complex numbers,which always has a solution $z = z_k / z_1$. Thus,the statement is False $(2)$.$(R)$ The values $z_1, z_2, \ldots, z_9$ are the roots of $\frac{z^{10}-1}{z-1} = 0$. Thus,$z^{10}-1 = (z-1)(z-z_1)(z-z_2)\ldots(z-z_9)$. Dividing by $(z-1)$,we get $1+z+z^2+\ldots+z^9 = (z-z_1)(z-z_2)\ldots(z-z_9)$. Setting $z=1$,we get $10 = (1-z_1)(1-z_2)\ldots(1-z_9)$. Taking the modulus,$|1-z_1||1-z_2|\ldots|1-z_9| = 10$. Thus,the expression equals $10/10 = 1$ $(3)$.$(S)$ The sum of all roots of $z^{10}-1=0$ is $1 + z_1 + z_2 + \ldots + z_9 = 0$. Thus,$\sum_{k=1}^9 z_k = -1$. Taking the real part,$\sum_{k=1}^9 \cos(2k\pi/10) = -1$. Then $1 - (-1) = 2$ $(4)$.Therefore,the correct matching is $P-1, Q-2, R-3, S-4$.

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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