The value of `Sigma_(i=1)^(6) (sin .(2pik)/(7)-i cos .(2pik)/(7))`

Options

A. -1
B. 0
C. `-i`
D. i

Answer

D

Explanation

Let `z=cos.(2pi)/(7)+ i sin. (2pi)/(7)`
Then by De Moivre's theorem, we have
`z^(k)=cos.(2pi k)/(7)+ i sin. (2 pi k)/(7)`
Now, `underset(i=1)overset(6)Sigma(sin .(2pi k)/(7)- (i cos 2 pi k)/(7))`
`=underset(i=1)overset(6)Sigma(-i)(cos.(2 pi k)/(7)+ i sin. (2k pi)/(7))`
`=(-i) underset(k=1)overset(6) Sigma z^(k)=-i(z(1-z^(6)))/(1-z)=-((z-z^(7))/(1-z))`
`=-(i)((z-1)/(1-z)) " " `[using `z^(7)=cos. 2 pi + i sin. 2 pi = 1`]
`=(-i)((1-z)/(1-z))=i`