Question

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

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

The correct option is C P-1, Q-2, R-3, S-4

(P)

$z_k$ is 10${}^{th}$ root of unity

$\Rightarrow {\overline{Z}}_k$ will also be 10${}^{th}$ root of unity. Take $z_j$ as ${\overline{z}}_k$.

(Q)

$z_1\ne 0$ take $z=\frac{z_k}{z_1}$, we can always find z.

(R)

$z^{10}-1=(z-1)(z-z_1)......(z-z_9)$
$\Rightarrow (z-z_1)(z-z_2).......(z-z_9)=1+z+z^2+......+z^9\forall z\in$ complex number.
Put $z=1$
$(1-z_1)(1-z_2).....(1-z_9)=10$

(S)

$1+z_1+z_2+.........+z_9=0$
$\Rightarrow Re\left(1\right)+Re\left(z_1\right)+.....+Re\left(z_9\right)=0$
$\Rightarrow Re\left(z_1\right)+Re\left(z_2\right)+......+Re\left(z_9\right)=-1$.
$\Rightarrow 1-{\sum }_{k=1}^{9}cos\frac{2k\pi }{10}=2$