Question

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
P	For each $z_{k}$ there exists a $z_{j}$ such $z_{k} \cdot z_{j}=1$	1	True
Q	There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1}, z=z_{k}$ has no solution $z$ in the set of complex numbers	2	False
R	$\frac{\left|1-z_{1}\right|\left|1-z_{2}\right| \ldots\left|1-z_{9}\right|}{10}$ equals	3	1
S	$1-\displaystyle\sum_{k=1}^{9} \cos \left(\frac{2 k \pi}{10}\right)$ 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) $z_{1} \neq 0$ take $z=\frac{z_{k}}{z_{1}}$, we can always find $z$.
(R) $z^{10}-1=(z-1)\left(z-z_{1}\right) \ldots\left(z-z_{9}\right)$
$\Rightarrow \left(z-z_{1}\right)\left(z-z_{2}\right) \ldots\left(z-z_{9}\right)=1+z+z^{2}+\ldots+z^{9} \forall z \in$ complex number
Put $z =1$
$\left(1-z_{1}\right)\left(1-z_{2}\right) \ldots\left(1-z_{9}\right)=10$
(S) $ 1+z_{1}+z_{2}+\ldots+z_{9}=0$
$\Rightarrow \text{Re}(1)+\text{Re}\left(z_{1}\right)+\ldots+\text{Re}\left(z_{9}\right)=0$ $\Rightarrow \text{Re}\left(z_{1}\right)+\text{Re}\left(z_{2}\right)+\ldots+\text{Re}\left(z_{9}\right)=-1$
$\Rightarrow 1-\displaystyle\sum_{k=1}^{9} \cos \frac{2 k \pi}{10}=2$