<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] As you say, if $a$ is a root, then $a\varepsilon$, where $\varepsilon = {1\over 2}+i{\sqrt{3}\over 2}$ is also a root. It is easy to see that $\varepsilon^2 = \varepsilon-1$. By Vieta we have $$a \varepsilon +a = -1\implies a = {-1\over \varepsilon+1}$$ so by second Vieta formula we have $$ b= 3a ^2\varepsilon = {3\varepsilon\over (\varepsilon+1)^2}={3\varepsilon\over \varepsilon^2+2\varepsilon+1} = {3\varepsilon\over \varepsilon-1+2\varepsilon+1}= 1$$
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