Roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle

Question

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!!

Answer 1

Obviously the roots shall be complex, so that $12b>9$.

$$ z_{1,2}=\frac{-3\pm i\sqrt{12b-9}}{6}\Rightarrow |z_1|=|z_2|=\sqrt{\frac{b}{3}},\quad |z_2-z_1|=\frac{\sqrt{12b-9}}{3}\\ \Rightarrow 3b=12b-9\Rightarrow b=1. $$

Answer 2

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 $b=1$

Answer 3

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 $$ b= 3a ^2\varepsilon = {3\varepsilon\over (\varepsilon+1)^2}={3\varepsilon\over \varepsilon^2+2\varepsilon+1} = {3\varepsilon\over \varepsilon-1+2\varepsilon+1}= 1$$