`z_(1),z_(2),z_(3)` are the vertices of an equilateral triangle taken in counter clockwise direction. If its circumcenter is at `(1-2i)` and `(z_(1)=2+i)`, then `z_(2)=`

Options

A. `(1-3sqrt(3))/2+(sqrt(3)-7)/(2)i`
B. `(1+3sqrt(3))/(2) -(7+sqrt(3))/(2)j`
C. `(1+3sqrt(3))/(2), (sqrt(3)-7)/(2)i`
D. `(1+3sqrt(3))/(2) +(7+sqrt(3))/(2)`i

Answer

A

Explanation

We have,
`bar(OA)` =Affix of A-Affix of O = `(2+i)-(1-2i)=1+3i`
Clearly, `bar(O)B` is obtained by rotating `bar(O)A` in anticlockwise sence through an angle `(2pi//3)`.

`therefore bar(OB)=bar(OA)e^(i2pi//3)`.
`rArr` (Affix of B- Affix of O) = `(1=3i)e^(i2pi//3)`
`rArr` Affix of B `-(1-2i)=(1+3i)(-1//2+i+isqrt(3)//2)`
`rArr` Affix of B `=1-2i+(-1//2-3sqrt(3)//2)+i(-3//2+sqrt(3)//2)`
`rArr` Affix of B =`(1-3sqrt(3))/2+i(sqrt(3)-7)/(2)`