Question

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

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

• A. P - (1) , Q - (2) , R - (4) , S - (3)
• B. P - (2) , Q - (1) , R - (3) , S - (4)
• C. P - (1) , Q - (2) , R - (3) , S - (4)
• D. P - (2) , Q - (1) , R - (4) , S - (3)

Answer 1

Answer: (C)

(P) $z_k$ is $10^{\text{th }}$ root of unity
$\Rightarrow {\bar{z}}_k$ will also be $10^{\text{th }}$ root of unity.
Take $z_j$ as ${\bar{z}}_k$ -
(Q) $z_1\ne 0$ take $z=\frac{z_k}{z_1}$, we can always find $z$.
(R) $z^{10}-1=(z-1)\left(z-z_1\right)\dots \left(z-z_9\right)$
$\Rightarrow \left(z-z_1\right)\left(z-z_2\right)\dots \left(z-z_9\right)=1+z+z^2+\dots +z^9\forall z\in$ complex number
Put $z=1$
$\left(1-z_1\right)\left(1-z_2\right)\dots \left(1-z_9\right)=10$
(S) $1+z_1+z_2+\dots +z_9=0$
$\Rightarrow \text{Re}(1)+\text{Re}\left(z_1\right)+\dots +\text{Re}\left(z_9\right)=0$ $\Rightarrow \text{Re}\left(z_1\right)+\text{Re}\left(z_2\right)+\dots +\text{Re}\left(z_9\right)=-1$
$\Rightarrow 1-\sum _{k=1}^9\cos \frac{2k\pi }{10}=2$