Let z k = cos ( 2 k 10 ) + i sin ( 2 k 10 ) , k = 1 , 2 , … , 9 . [2014]

Question

$Let z_{k} = \cos (\frac{2 k π}{10}) + i \sin (\frac{2 k π}{10}), k = 1, 2, \dots, 9 .$ [2014]

	List - I		List - II
P.	For each $z_{k}$ there exists $z_{j}$ such that $z_{k} . z_{j} = 1$	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.	$1 - \sum_{k = 1}^{9} \cos (\frac{2 k π}{10})$ equals	4.	2

Smart Achievers · Accepted Answer

(3)

$(P) \to (1) : z_{k} = \cos \frac{2 k π}{10} + i \sin \frac{2 k π}{10}, k = 1 to 9$

$∴$ $z_{k} = e^{i \frac{2 k π}{10}}$

$Now z_{k} . z_{j} = 1 \Rightarrow z_{j} = \frac{1}{z_{k}} = e^{- i \frac{2 k π}{10}} = \bar{z_{k}}$

$We know if z_{k} is 10th root of unity so will be \bar{z_{k}}$

$∴$ $For every z_{k} there exist z_{i} = \bar{z_{k}}$

$Such that z_{k} . z_{j} = z_{k} . \bar{z_{k}} = 1$

Hence the statement is true.

$(Q) \to (2) : z_{1} = z_{k} \Rightarrow z = \frac{z_{k}}{z_{1}}, z_{1} \neq 0$

$∴$ $We can always find a solution of z_{1} z = z_{k}$

Hence the statement is false.

$(R) \to (3) : We know z^{10} - 1 = (z - 1) (z - z_{1}) \dots (z - z_{9})$

$\Rightarrow (z - z_{1}) (z - z_{2}) \dots (z - z_{9}) = \frac{z^{10} - 1}{z - 1} = 1 + z + z^{2} + \dots + z^{9}$

$For z = 1, we get (1 - z_{1}) (1 - z_{2}) \dots (1 - z_{9}) = 10$

$∴$ $\frac{| 1 - z_{1} | | 1 - z_{2} | \dots | 1 - z_{9} |}{10} = 1$

$(S) \to (4) : 1, z_{1}, z_{2}, \dots, z_{9} are 10th roots of unity.$

$∴$ $z^{10} - 1 = 0$

$From equation 1 + z_{1} + z_{2} + \dots + z_{9} = 0,$

$Re (1) + Re (z_{1}) + Re (z_{2}) + \dots + Re (z_{9}) = 0$

$\Rightarrow Re (z_{1}) + Re (z_{2}) + \dots + Re (z_{9}) = - 1$

$\Rightarrow \sum_{k = 1}^{9} \cos \frac{2 k π}{10} = - 1$ $\Rightarrow 1 - \sum_{k = 1}^{9} \cos \frac{2 k π}{10} = 2$

$Hence (3) is the correct option.$

Want Us to Email you About Special Offers & Updates?