Question

If $z_1,z_2$ are the roots of the equation $a{z}^2+bz+c=0$, with $a,b,c>0;2b^2>4ac>b^2;z_1,\in$ third quadrant ; $z_2\in$ second quadrant in the argand's plane then, show that
$arg\left(\frac{z_1}{z_2}\right)=co{s}^{-1}{\left(\frac{b^2}{4ac}\right)}^{1/2}$

Answer 1

$b^2-4ac<0$ hence complex.

$z_1=\frac{-b}{2ac}+\frac{i\sqrt{4ac-b^2}}{2ac}$

$z_2=\frac{-b}{2ac}-\frac{i\sqrt{4ac-b^2}}{2ac}$

$\frac{z_1}{z_2}=\frac{z_1\overline{z_2}}{|z_2{|}^2}$

$\frac{{\left(\frac{-b+i\sqrt{4ac-b^2}}{2ac}\right)}^2}{\left(\frac{{-b}^2+4ac-b^2}{{4\left(ac\right)}^2}\right)}=\frac{\left(\frac{b^2-4ac+b^2}{{4\left(ac\right)}^2}\right)-\frac{i\sqrt{4ac-b}}{2c{a}^2}}{\frac{-1}{ac}}⟹\frac{2b^2-4ac}{4\left(ac\right)}-\frac{i\sqrt{4ac-b^2}}{2ca}$

$arg\left(\frac{z_1}{z_2}\right)={\tan }^{-1}\left[\frac{i\sqrt{4ac-b^2}}{2ca}\times \frac{4\left(ac\right)}{2b^2-4ac}\right]={\cos }^{-1}\sqrt{\frac{b^2}{4ac}}$