If `|z_(1)|=|z_(2)|` and arg `(z_(1))+"arg"(z_(2))=0`, then

Options

A. `(7sqrt(7))/(2sqrt(3))`
B. `(5sqrt(7))/(2sqrt(3))`
C. `(14sqrt(7))/(sqrt(3))`
D. `(7sqrt(7))/(5sqrt(3))`

Answer

B

Explanation

We have,
`|sqrt(3)(1-2z)+2i|=2sqrt(7)`
`|sqrt(3)(1-2z)+2i|=2sqrt(7)`
`rArr |-2sqrt(3z)+(sqrt(3)+2i)|=2sqrt(7)`
`rArr |z-(1/2+1/sqrt(3)i)|=sqrt(7/3)`
Clearly, it represents a circle haivng center at `(1//2,1//sqrt(3))` and radius `r_(1)=sqrt(7/3)`. it is given that `z_(1)` lies on (i)

The equation of another curve is
`|sqrt(3)(-1-z)-2i|=|sqrt(3)(9-z)+18i|`
or `|-1-z-2/sqrt(3)i|=|9-z+6sqrt(3)i|`
or `,|z+1+2/sqrt(3)i|=|z-9-6sqrt(3)i|`
or `,|z-(-1-2/sqrt(3)i)|=|z-(9+6sqrt(3)i)|`
This, represents perpendicular bisector of the line segement joining point `A(-1,-2/sqrt(3))` and `B(9,6sqrt(3))`. The coordinates of the mid-point C of AB are `(4,8//sqrt(3))`. Clearly, A,B and the center of the circle are collinear.