Complex Numbers Part 1

Given quadratic equation, $Z^2+a Z+b=0$ and two roots are $Z_1$ and $Z_2$. $\therefore Z_1+Z_2=-a$ and $Z_1 Z_2=b$ Question says,There are three complex numbers: 1. Origin $(0)$ 2. $Z_1$ 3. $Z_2$ and they form an equilateral triangle. So They are the vertices of the triangle. [ Important Point : If $Z_1, Z_2$ and $Z_3$ are the vertices of an equilateral triangle then -$Z_1^2+Z_2^2+Z_3^2=Z_1 Z_2+Z_2 Z_3+Z_3 Z_1]$ In this question,$Z_1=0, Z_2=Z_1 \text{ and } Z_3=Z_2$ By putting those values in the equation we get, $0^2+Z_1^2+Z_2^2=0+Z_1 Z_2+0$ $\Rightarrow Z_1^2+Z_2^2=Z_1 Z_2$$\Rightarrow Z_1^2+Z_2^2=b [.$ as $Z_1 Z_2=b]$$\Rightarrow (Z_1+Z_2)^2-2 Z_1 Z_2=b$$\Rightarrow (Z_1+Z_2)^2-2b=b$$\Rightarrow (Z_1+Z_2)^2=3b$$\Rightarrow (-a)^2=3b$$\Rightarrow a^2=3b$ So Option (D) is correct.[ Note : This question is asked to check if you know the following formula - "If $Z_1, Z_2$ and $Z_3$ are the vertices of an equilateral triangle then -$Z_1^2+Z_2^2+Z_3^2=Z_1 Z_2+Z_2 Z_3+Z_3 Z_1{}^{''}]$