Let z k = cos ⁡ 2 k π 10 + i sin ⁡ 2 k π 10 ; k = 1 ,   2 , … , 9 List – I List – II (A) For each z k there exists a z j   s u c h   z k   .   z j = 1 (P) True (B) There exists a   k   ϵ   1 ,   2 ,  …

Question

Let z k = cos &#8289; 2 k &#960; 10 + i sin &#8289; 2 k &#960; 10 ; k = 1 , &#160; 2 , &#8230; , 9 List &ndash; I List &ndash; II (A) For each z k there exists a z j &#160; s u c h &#160; z k &#160; . &#160; z j = 1 (P) True (B) There exists a &#160; k &#160; &#1013; &#160; 1 , &#160; 2 , &#160; &#8230; , 9 such that z 1 &#160; . &#160; z = z k has no solution z in the set of complex numbers (Q) False (C) 1 - z 1 &#160; 1 - z 2 &#8230; 1 - z 9 &#160; 10 &#160; equals (R) 1 (D) 1 - &#8721; k = 1 9 cos &#8289; 2 k &#960; 10 &#160; equals (S) 2

Aicharya · Accepted Answer

P z k i s 10 t h root of unity ⇒ z k - will also be 10 t h root of unity. Take z j as z k - Q z 1 ≠ 0 t a k e z = z k z 1 , we can always find z. R z 10 - 1 = z - 1 z - z 1 … z - z 9 ⇒ z - z 1 z - z 2 … z - z 9 = 1 + z + z 2 + … + z 9 ∀ z ϵ complex number. Put z = 1 1 - z 1 1 - z 2 … 1 - z 9 = 10 S 1 + z 1 + z 2 + … + z 9 = 0 ⇒ R e 1 + R e z 1 + … + R e z 9 = 0 ⇒ R e z 1 + R e z 2 + … + R e z 9 = - 1 ⇒ 1 - ∑ k = 1 9 cos ⁡ 2 k π 10 = 2

Let z k = cos ⁡ 2 k π 10 + i sin ⁡ 2 k π 10 ; k = 1 ,   2 , … , 9 List…

Solution

	List – I		List – II
(A)	For each $z_{k}$ there exists a $z_{j} s u c h z_{k} . z_{j} = 1$	(P)	True
(B)	There exists $a k ϵ \{1, 2, \dots, 9\}$ such that $z_{1} . z = z_{k}$ has no solution z in the set of complex numbers	(Q)	False
(C)	$\frac{\|1 - z_{1}\| \|1 - z_{2}\| \dots \|1 - z_{9}\|}{10}$ equals	(R)	1
(D)	$1 - \sum_{k = 1}^{9} \cos (\frac{2 k π}{10})$ equals	(S)	2

Let z k = cos &#8289; 2 k &#960; 10 + i sin &#8289; 2 k &#960; 10 ; k = 1 , &#160; 2 , &#8230; , 9 List…

Solution

Let z k = cos ⁡ 2 k π 10 + i sin ⁡ 2 k π 10 ; k = 1 , 2 , … , 9 List…