Question

Let origin and the non-real roots of $2z^2+2z+\lambda =0$ form the three vertices of an equilateral triangle in the Argand plane. Then $3\lambda =$

Answer 1

$2z^2+2z+\lambda =0$
Let the roots be $z_1,z_2$
$z_1+z_2=-1,z_1{z}_2=\frac{\lambda }{2}$

$\because 0,z_1,z_2$ form an equilateral triangle,
$\therefore z_1=z_2{e}^{i\pi /3}$ (using rotation)
$\Rightarrow -1-z_2=-z_2{\omega }^2[\because e^{i\pi /3}=-{\omega }^2]$
$\Rightarrow z_2=\frac{-1}{1-{\omega }^2}$

$\therefore \lambda =2z_2(-{\omega }^2)z_2=-2{\omega }^2{z}_2^2=\frac{-2{\omega }^2}{1+\omega -2{\omega }^2}=\frac{2}{3}[\because {\omega }^3=1,1+\omega +{\omega }^2=0]\Rightarrow 3\lambda =2$