<s>[INST] Roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle Given that roots of $3z^2+3z+b$  and $0$ lie on equilateral triangle, find condition on $b$.
If $a$ is a  root then $a(\frac{1}{2}+i\frac{\sqrt 3 }{2})$ i also a root. Sum of root is $-1 = a(\frac{3}{2}+i\frac{\sqrt3}{2})$ and product is $a^2(\frac{1}{2}+i\frac{\sqrt 3 }{2}) = \frac{b}{3}$
I am asking is this correct and also does veeta's formula work in complex numbers? What is your method and how to proceed using this, find $a$ and put in second equation to get $b$ is complicated.
Thanks  a lot!!
 [/INST] Let the roots of the quadratic be $p\pm iq$
Looking at the sum of roots, we have $p=-\frac 12$
Since the triangle is equilateral, $$\sqrt{p^2+q^2}=2|q|$$
So $$q=\frac{1}{2\sqrt{3}}$$
Looking at the product of roots, $$\frac b3=p^2+q^2=\frac 13$$
Hence $b=1$
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