Question

If a and b are negative real numbers such that the points $z_1=a-i,z_2=-1+ib$ and $z_3=0$ form an equilateral triangle; then the pair (a, b) is equal to ................ & ..............

• A. (2 + $\sqrt{3}$, 2 + $\sqrt{3}$), (2 -$\sqrt{3}$, 2- $\sqrt{3}$)
• B. (-2 + 2$\sqrt{3}$, -2 + 2$\sqrt{3}$), (-2 -2$\sqrt{3}$, -2- 2$\sqrt{3}$)
• C. (2 +2 $\sqrt{3}$, 2 +2 $\sqrt{3}$), (2 -2$\sqrt{3}$, 2- 2$\sqrt{3}$)
• D. (-2 + $\sqrt{3}$, -2 + $\sqrt{3}$), (-2 -$\sqrt{3}$, -2- $\sqrt{3}$)

Answer 1

Answer: (D)

(-2 + $\sqrt{3}$, -2 + $\sqrt{3}$), (-2 -$\sqrt{3}$, -2- $\sqrt{3}$)

Hence
$|z_3-z_1|=|z_3-z_2|$
$a^2+1=b^2+1$
Or
$a=\pm b$
and
$z_1^2+z_2^2+z_3^2=z_1.z_2+z_3{z}_2+z_3{z}_1$
$a^2-1-2ai+1-b^2-2bi=(a-b)+i(1+ab)$
$(a^2-b^2)-i\left(2\right(a+b\left)\right)=-(a-b)+i(1+ab)$
Hence
$-2(a+b)=1+ab$
Considering $b=a$, we get
$b^2+1+4b=0$
$b=\frac{-4\pm \sqrt{16-4}}{2}$
$=-2\pm \sqrt{3}$