Complexity!

$z_1,z_2,z_3$ form an equilateral triangle implies

$z^2_1+z^2_2+z^2_3= z_1z_2 + z_2z_3+z_3z_1.$

Here $z_3=0$ .

$\Rightarrow z^2_1+z^2_2=z_1z_2$

$\Rightarrow a^2-2b=b$

Hence we have $\boxed{a^2=3b}$ .

$\begin{aligned} z^2+az+b & = 0 \quad \Rightarrow z = \frac{-a \pm \sqrt{a^2-4b}}{2} \quad \Rightarrow \begin{cases} z_1 = -\frac{a}{2} + i \frac{\sqrt{4b-a^2}}{2} \\ z_2 = -\frac{a}{2} - i \frac{\sqrt{4b-a^2}}{2} \end{cases} \end{aligned}$

For equilateral triangle:

$\begin{aligned} \Rightarrow |z_1 - z_2| & = |z_1 - 0| = |z_2-0| \\ |z_1 - z_2|^2 & = |z_1 - 0|^2 = |z_2-0|^2 \end{aligned}$

$\begin{aligned} \small \left| \left( -\frac{a}{2} + i \frac{\sqrt{4b-a^2}}{2} \right) - \left(- \frac{a}{2} - i \frac{\sqrt{4b-a^2}}{2} \right) \right|^2 & = \small \left| \left( -\frac{a}{2} \color{#D61F06}{\pm} i \frac{\sqrt{4b-a^2}}{2} \right) - \left(0 \right) \right|^2 \\ 4\left( \frac{4b-a^2}{4} \right) & = \frac{a^2}{4} + \frac{4b-a^2}{4} \\ 3(4b-a^2) & = a^2 \\ 12b & = 4a^2 \quad \Rightarrow \boxed{a^2 = 3b} \end{aligned}$

It's given that $z_1,z_2$ are the two complex roots (real/non-real) of the equation $z^2+az+b=0$ . We're also given that $a,b\in\Bbb R$ . By Vieta's Formula , we have,

$a=-(z_1+z_2)\in\Bbb R~,~b=z_1z_2\in\Bbb R~~\ldots~(i)$

Now, if one of the roots is real and the other one is non-real, then obviously $a$ cannot be real. For $a$ to be real, we need the imaginary parts of $z_1$ and $z_2$ to be additive inverses of each other so that the resultant imaginary part in the sum of the roots becomes zero. Similar argument goes for $b$ to be real since product of one real and another non-real value can never be real. For both the sum and product of the roots to be real, the roots must be complex conjugates. Hence, we have,

$a,b\in\Bbb R\iff z_1=\overline{z_2}~~,~z_1,z_2\in\Bbb C~~\ldots~(ii)$

Let $x=\Re(z_1)$ and $y=\Im(z_1)$ . Using $(ii)$ , we have $z_2=x-iy$ .

Now, we're also given that the origin, $z_1$ and $z_2$ form an equilateral triangle in the Argand plane (complex plane). Recall the distance formula used to calculate the distance between two points. Using that and the equilateral triangle condition given (all sides equal), we get,

$|z_1|=|z_2|=|z_2-z_1|$

Square the above equation and use $(ii)$ to obtain,

$x^2+y^2=(2y)^2=4y^2\implies x^2=3y^2~~\ldots~(iii)$

Use $(i)$ and $(ii)$ to obtain,

$a=-2x\implies a^2=4x^2~;~b=x^2+y^2~~\ldots~(iv)$

Use $(iii)$ and $(iv)$ to obtain,

$a^2=4x^2=4\times 3\times y^2=3\times b=3b$

$\boxed{\therefore\quad a^2=3b}$

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Clarifications:

• $\Re(z)$ is used to denote the real part of a complex number $z$ .
• $\Im(z)$ is used to denote the imaginary part of a complex number $z$ .
• $i$ stands for the imaginary unit, which is $i=\sqrt{-1}$ .
• $|z|$ denotes the absolute value (modulus) of a complex number $z$ .
• $\Bbb R$ and $\Bbb C$ are used to denote the set of reals and set of complex numbers respectively.

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By the way, in case anyone forgot the distance formula, here it is:

The distance between two points $(a,b)$ and $(c,d)$ is given by,

$\textrm{Distance}=\sqrt{(c-a)^2+(d-b)^2}=\bigg|(c+id)-(a+ib)\bigg|$

Since the given quadratic equation has real coefficients, it follows by the Complex Conjugate Root Theorem that the two complex solutions must be conjugates, which in turn means that they are reflections of each other over the real axis. Thus in order to form an equilateral triangle with the origin, the two solutions must make angles of $\frac{\pi}{6}$ (one above, the other below) with either the positive or negative half of the real axis.

Therefore, for some $r\not=0$ , we may write

$z_1 = r\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right); \quad z_2 = r\left(\cos\dfrac{\pi}{6} - i\sin\dfrac{\pi}{6}\right)$

$z_1 = \dfrac{r}{2}(\sqrt{3} + i); \quad z_2 = \dfrac{r}{2}(\sqrt{3} - i)$

Note that the solutions on the negative side of the real axis are accounted for with any $r<0$ .

Then applying Vieta's Formulas , we get

$-a = z_1 + z_2 = \dfrac{r}{2}(2\sqrt{3}) = r\sqrt{3}$

and

$b = z_1z_2 = \dfrac{r^2}{4}(3 - i^2) = r^2$

which then yields

$a^2 = (r\sqrt{3})^2 = 3r^2 = 3b$ .