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$z_1$ and $z_2$ are the roots of $3z^2+3z+b=0$.If $O(0),A(z_1),B(z_2)$ is an equilateral triangle then what will be the value of b ?

My approach:I took $z_1=m_1+in_1$ and $z_2=m_2+in_2$ and proceeded like a normal quadratic finding sum and product of roots.But after that what should i do?How should i proceed?

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Your approach is basically good, but it will be much easier if you use some polar forms. From a diagram you can see that if $OAB$ is equilateral then $$z_2=z_1e^{\pm\pi i/3}\ .$$ The sum of roots gives $z_1+z_2=-1$, that is, $$z_1(1+e^{\pm\pi i/3})=-1\ .$$ Now using the product of roots, $$b=3z_1z_2=\frac{3e^{\pm\pi i/3}}{(1+e^{\pm\pi i/3})^2}\ .$$ Doing the complex arithmetic carefully shows that both the $+$ sign and the $-$ sign give the same answer, $b=1$.


Edit: calculation details provided by the OP.

Case 1: $$b=3z_1z_2=3(-ω^2)/(1-ω^2)^2=(-3ω^2)/(1+ω-2ω^2)=(-3ω^2)/(-3ω^2)=1$$ Case 2: $$b=3z_1z_2=3(-ω)/(1-ω)^2=(-3ω)/(1+ω^2-2ω)=(-3ω)/(-3ω)=1$$

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