Let 
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 
{1, 2, … .., 9} such that
z 1 . z = z k has no solution z in the set of
complex numbers (2) False
(R) 
equals (3) 1
(S) 
equals (4) 2
Codes:
P Q R S
Text Solution
Verified by ExpertsC
(P) z k is 10 th root of unity 
will also be 10 th root of unity. Take z j as 
.
(Q) z 1 
0 take 
, 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 
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.
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