Let `z,w` be complex numbers such that `barz+ibarw=0` and `arg zw=pi` Then `argz` equals

Options

A. 4
B. 6
C. 8
D. more than 8

Answer

C

Explanation

Let `z=cosx+isinx,x in[0,2pi)`. Then,
`1=|(z)/(barz)+(barz)/(z)|`
`=(|z^(2)+barz^(2)|)/(|z|^(2))`
`=|cos2x+isin2x+cos2x-isin2x|`
`=2|cos2x|`
`rArrcos2x =pm1//2`
Now, `cos2x=1//2`
`rArr x_(1)=(pi)/(6),x_(2)=(5pi)/(6),x_(3)=(7pi)/(6),x_(4)=(11pi)/(6)`
`cos2x=-(1)/(2)`
`rArr x_(5)=(pi)/(3),x_(6)=(2pi)/(3),x_(7)=(4pi)/(3),x_(8)=(5pi)/(3)`