Question

show that the origin and the complex numbers represented by the roots of the equation z^2+az+b=0 form an equilateral triangle if a^2=3b

Answer 1

Answer:

Let the root of the equation $z^2+az+b=0$ be $z_1$ and  $z_2$

Then

$z_1+z_2=-a$      (sum of roots)

and $z_1z_2=b$    (product of roots)

If (0,0), $z_1$ and  $z_2$ form the vertices of an equilateral triangle

It is worth remembering that if

z₁, z₂ and z₃ represent the vertices of an equilateral triangle

then

z₁²+z₂²+z₃²=z₁z₂+z₂z₃+z₃z₁

or, $z_1^2+z_2^2+0^2=z_1z_2+z_2\times 0+z_1\times 0$

$\implies z_1^2+z_2^2=z_1z_2$

$\implies z_1^2+z_2^2+2z_1z_2=z_1z_2+2z_1z_2$

$\implies (z_1+z_2)^2=3z_1z_2$

$\implies (-a)^2=3b$

$\implies a^2=3b$