If is a 6th root of unity, then

1.

2. 

3. 

4.

Hello dear

ans : option (1)

let   6 = 1

then (-1 ) ( 1++2+3+4+5) = 0

( 1++2+3+4+5) = 0

( 1++2+3+4)= - 5 = - ( cos 5pi/3 + i sin 5pi/3)

! ( 1++2+3+4)! = ! - ( cos 5pi/3 + i sin 5pi/3)! = 1

so option (1) is right one.

This was in BMAT..

Solution:

1,3 = 1 × 1 [Modulus of each 6th root of unity is 1]