Q:

Roots(Complex Number) Arguments

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
$$\arg\dfrac{\ z_1}{\ z_2}= 2\arccos\Bigl( \dfrac{b^2}{4ac}\Bigr)^{1/2}$$
Concept Used $\ z_1 + \ z_2 = -\dfrac{b}{a}$
As $b^2<4ac$, the roots are complex, 
$u,v,w,x>0, \quad z_1=-u-iv\;$  & $\; z_2=-w+ix$.  From here, I am not able to proceed. 

A:

Since $b^2<4ac$, we have that
$$z_2=-\frac{b}{2a}+i\frac{\sqrt{4ac-b^2}}{2a}=\overline{z_1}.$$
Therefore
$$\frac{z_1}{z_2}=\frac{z_1\overline{z_2}}{|z_2|^2}=
\frac{z_1^2}{|z_1|^2}\implies \mbox{arg}(\frac{z_1}{z_2})=\mbox{arg}(z_1^2)=2\mbox{arg}(z_1).$$
Can you take it from here?