Complex roots forming a equilateral triangle

Question

Suppose we have relation $$z^2 + az + b=0 $$ where $a$ and $b$ are real and roots of this equation $z_1$ and $z_2$ form equilateral triangle with origin then what could be relation between $a$ and $b$ ? I simply applied quadratic formula in equation $$ z = \frac{-a \pm \sqrt{a^2 -4b}}{2} $$ now since it forms equilateral triangle with origin so $$|z_1| =|z_2|$$ applying which $$( \frac{-a + \sqrt{a^2 -4b}}{2})^2= (\frac{-a - \sqrt{a^2 -4b}}{2})^2$$ I arrived at $$a^2 = 4b$$ but my answer is incorrect , why?

Answer 1

Hint :

$z_1,z_2$ are complex numbers $$(z-z_1)(z-z_2)=z^2+az+b$$ $$z_1+z_2=-a$$

$$z_1z_2=b$$

Because of equal triangle, We know that $$z_1=M e^{i \alpha }$$ $$z_2=M e^{i (\alpha+ \pi /3) }$$ Thus you can use a relation between two complex numbers.

$$z_2/z_1=e^{i \pi /3}= \frac{ 1}{2}+i \frac{ \sqrt3}{2} $$

Answer 2

Hint; The triangle in the complex plane whose verticies are the origin and the points $z_{1}$ and $z_{2}$ is equalateral if and only if \begin{equation} z_{1}+z_{2} = z_{1}z_{2} \end{equation} (Further hint: The reasoning behind this is that the distance from the origin to $z_{1}$ is $\sqrt{z_{1}\bar{z_{1}}}$ etc...)