Let z _ k = ((2kπ)/(10)) + isin ((2kπ)/(10)); k = 1, 2, , 9 List – I List – II (A) For each z _ k there exists a z _ j such z _ k . z _ j = 1 (P) True (B) There exists akϵ\ 1, 2, , 9\ such that z _ 1 . z = z _ k has no

Correct answer: B. a-p;b-q;c-r;d-s; (P) z _ k is 10 ^ th root of unity ⇒ - z _ k will also be 10 ^ th root of unity. Take z _ j as - z _ k (Q) z _ 1 ≠ 0takez = 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 zϵ complex number. Put z = 1 (1 - z _ 1 )(1 - z _ 2 ) (1 - z _ 9 ) = 10 (S)1 + z _ 1 + z _ 2 + + z _ 9 = 0 ⇒ Re(1) + Re( z _ 1 ) + + Re( z _ 9 ) = 0 ⇒ Re( z _ 1 ) + Re( z _ 2 ) + + Re( z _ 9 ) = - 1 ⇒ 1 - ∑ _ k = 1 ^ 9 cos (2kπ)/(10) = 2

JEE Advanced 2014Paper 2

Exam: JEE Advanced
Year: 2014
Paper: Paper 2
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2014 · Paper 2 · Mathematics · Complex Number

Let z _ k = ((2kπ)/(10)) + isin ((2kπ)/(10)); k = 1, 2, , 9 List –...

Let

{z}_{k} = \cos \left(\dfrac{2k\pi}{10}\right) + isin \left(\dfrac{2k\pi}{10}\right); k = 1, 2, \ldots , 9

	List – I		List – II
(A)	For each ${z}_{k}$ there exists a ${z}_{j}such{z}_{k} . {z}_{j} = 1$	(P)	True
(B)	There exists $akϵ\left\{1, 2, \ldots , 9\right\}$ such that ${z}_{1} . z = {z}_{k}$ has no solution z in the set of complex numbers	(Q)	False
(C)	$\dfrac{\left\|1 - {z}_{1}\right\|\left\|1 - {z}_{2}\right\| \ldots \left\|1 - {z}_{9}\right\|}{10}$ equals	(R)	1
(D)	$1 - {∑}_{k = 1}^{9}\cos \left(\dfrac{2k\pi}{10}\right)$ equals	(S)	2

Options

Aa-p;b-s;c-r;d-q;
Ba-p;b-q;c-r;d-s;
Ca-r;b-p;c-q;d-s;
Da-p;b-r;c-s;d-q;

Correct answer

B. a-p;b-q;c-r;d-s;

Step-by-step solution

\left(P\right){z}_{k}is{10}^{th}

root of unity

⇒ \overset{-}{{z}_{k}}

will also be

{10}^{th}

root of unity. Take

{z}_{j}

\overset{-}{{z}_{k}}

\left(Q\right){z}_{1} \ne 0takez = \dfrac{{z}_{k}}{{z}_{1}},

we can always find z.<br />

\left(R\right){z}^{10} - 1 =

\left(z - 1\right) \left(z - {z}_{1}\right) \ldots \left(z - {z}_{9}\right)<br />

⇒ \left(z - {z}_{1}\right)\left(z - {z}_{2}\right)

\ldots \left(z - {z}_{9}\right) = 1 + z + {z}^{2} + \ldots + {z}^{9} \forall zϵ

complex number.
Put z = 1
\left(1 - {z}_{1}\right)\left(1 - {z}_{2}\right) \ldots \left(1 - {z}_{9}\right) = 10

<br />\left(S\right)1 + {z}_{1} + {z}_{2} + \ldots + {z}_{9} = 0

⇒ Re\left(1\right) + Re\left({z}_{1}\right) + \ldots + Re\left({z}_{9}\right) = 0

⇒ Re\left({z}_{1}\right) + Re\left({z}_{2}\right) + \ldots + Re\left({z}_{9}\right) = - 1

⇒ 1 - \mathop{∑}_{k = 1}^{9}cos \dfrac{2k\pi}{10} =

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