Let z_{k} = cos \textrm{ } \left(\right. \frac{2 kπ}{10} \left.\right) + isin \left(\right. \frac{2 kπ}{10} \left.\right); k = 1, 2, …….,9. List I List II (P) For each zk there exists a zj such that zk · zj = 1 (1) True (Q) There exists a k \in {1, 2, ……., 9} such that z1 · z = zk has no solution z in the set of complex numbers (2) False (R) \frac{\left|\right. 1 - z_{1} \left|\right. \textrm{ } \left|\right. 1 - z_{2} \left|\right. \textrm{ } \ldots \ldots . \textrm{ } \left|\right. 1 - z_{9} \left|\right.}{10} equals (3) 1 (S) 1 - \sum_{k = 1}^{9} cos \textrm{ } \left(\right. \frac{2 kπ}{10} \left.\right) equals (4) 2

Engineering

Mathematics

nth Roots of Unity

Question

Let $z_{k} = \cos (\frac{2 kπ}{10}) + isin (\frac{2 kπ}{10})$ ; k = 1, 2, …….,9.

List I	List II
(P) For each z_k there exists a z_j such that z_k · z_j = 1	(1) True
(Q) There exists a k $\in$ {1, 2, ……., 9} such that z₁ · z = z_k has no solution z in the set of complex numbers	(2) False
(R) $\frac{\| 1 - z_{1} \| \| 1 - z_{2} \| \dots \dots . \| 1 - z_{9} \|}{10}$ equals	(3) 1
(S) $1 - \sum_{k = 1}^{9} \cos (\frac{2 kπ}{10})$ equals	(4) 2

P → 2

Q → 1

R → 3

S → 4

P → 1

Q → 2

R → 3

S → 4

P → 1

Q → 2

R → 4

S → 3

P → 2

Q → 1

R → 4

S → 3

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z¹⁰ = 1 where z $\neq$ 1

(P) True

(S) $1 - \sum_{k = 1}^{9} \cos (\frac{2 πk}{10})$ = 1 – (–1) = 2.