Let \(z_k=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right) ; k=1,2, \ldots, 9\)

List - I	List - II
(A) For each \(z_k\) there exists a \(z_j\) such \(z_k \cdot z_j=1\)	(P) True
(B) 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	(Q) False
(C) \(\frac{\left|1-z_1\right|\left|1-z_2\right|\ldots\left|1-z_9\right|}{10}\) equals	(R) 1
(D) \(1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)\) equals	(S) 2

(P) \(z_k\) is \(10^{\text {th }}\) root of unity \(\Rightarrow \overline{z_k}\) will also be \(10^{\text {th }}\) root of unity. Take \(z_j\) as \(\overline{z_k}\)
(Q) \(z_1 \neq 0\) take \(z=\frac{z_k}{z_1}\), we can always find z .
\((R) z^{10}-1=(z-1) \left(z-z_1\right) \ldots\left(z-z_9\right)\)
\(\Rightarrow \left(z-z_1\right) \left(z-z_2\right) \ldots\left(z-z_9\right)=1+z+z^2+\ldots\) \(+~z^9 \forall z \epsilon\) complex number.
Put z = 1
\(\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_9\right)=10 \)
\( (S) \quad 1+z_1+z_2+\ldots+z_9=0 \)
\( \Rightarrow \operatorname{Re}(1)+\operatorname{Re}\left(z_1\right)+\ldots+\operatorname{Re}\left(z_9\right)=0 \)
\( \Rightarrow \operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)+\ldots+\operatorname{Re}\left(z_9\right)=-1 \)
\( \Rightarrow 1-\sum_{k=1}^9 \cos \frac{2 k \pi}{10}=2\)