If a and b are two real number lying between 0 and 1 such that `z_1=a+i, z_2=1+bi and z_3=0` form anequilateral trilangle , then

Options

A. `a=2+sqrt(3)`
B. `b=4-sqrt(3)`
C. `a=b=2-sqrt(3)`
D. `a=2,b=sqrt(3)`

Answer

C

Explanation

Since `z_1,z_2 "and " z_3 ` from an equilateral triangle
`rArr " " z_1^2+z_2^2+z_3^2=z_1z_2+z_2^3+Z_3z_1`
`rArr " " (a+i)^2 +(1+ib)^2+(0)^2=(a + i)(1+ib)+0+0`
`rArr " "a^2-1 + 2ia + 1 -b^2+2ib =a + I (ab + 1 )-b `
`rArr (a^2 -b^2)+2i (a+b)= (a-b)+i (ab+1)`
`rArr " "a^2-b^2=a-b`
and `" "2(a+b)=ab + 1`
If `a=b a(2a)=a^2 + 1`
`rArr a^2 - 4 a +1 =0 `
`rArr " " a^2-4a+1=0`
`rArr a= (4 pmsqrt(16-4))/2=2 pm sqrt(3)`
If `a+b = 1,2 2 =a (1-a)+1 rArr a^2-a + 1=0 `
`rArr a=(1 pm sqrt(1-4))/(2)` but a and b `b in R`
`therefore` Only solution when a=b
`rArr " " a=b=2 pi sqrt(3)`
`rArr " " a=b= 2-sqrt (3)" [ because a, b in (0,1)]`