$\arg(\frac{z_1}{z_2})$ of complex equation If $z_1,z_2$ are the roots of the equation $az^2 + bz + c = 0$, 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 $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$

Since $a,b,c$ are real then the roots are $\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}$.
Given the quadrants locations, $z_1$ is the root with $-$ sign.
Thus $\dfrac{z_1}{z_2}=\dfrac{-b-i\sqrt{4ac-b^2}}{-b+i\sqrt{4ac-b^2}}=\dfrac{(b+i\sqrt{4ac-b^2})^2}{4ac}=\dfrac{(b^2-2ac)+ib\sqrt{4ac-b^2}}{2ac}$
It is quite easy to see that this complex is of module $1$.
[ $\require{cancel}(b^2-2ac)^2+b^2(4ac-b^2)=\cancel{b^4}-\cancel{4b^2ac}+4a^2c^2+\cancel{4b^2ac}-\cancel{b^4}=4a^2c^2$ ]
So $\cos(\theta)=\dfrac{b^2-2ac}{2ac}=\dfrac{b^2}{2ac}-1\iff 1+cos(\theta)=2\cos(\frac{\theta}2)^2=\dfrac{b^2}{2ac}$
Given the conditions of positivity for $a,b,c$ and $b^2-2ac$, we get that $\theta$ is in the first quadrant and the formula given in the problem.
