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 $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$

  • $\begingroup$ arg is multivalued or set valued and the same for $\_{cos}^{-1}$ $\endgroup$ – StuartMN Jan 28 '18 at 4:57

$z_2 = \overline{z_1}\,$ since the quadratic has real coefficients, so $\,\arg\left(\dfrac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)=2 \arg(z_1)\,$.

Since $\,a,b,c \gt 0\,$ the roots are in the left half-plane $\,\operatorname{Re}(z_1)=\operatorname{Re}(z_2) = -\,\dfrac{b}{2a} \lt 0\,$, and given the condition that "$z_1\in$ third quadrant" $\,\operatorname{Im}(z_1) \lt 0\,$, so $\,z_1\,$ is the root with negative imaginary part:

$$z_1 = \dfrac{-b - i \sqrt{4ac - b^2}}{2a} = \sqrt{\dfrac{c}{a}}\left(-\dfrac{b}{2\sqrt{ac}} - i \sqrt{1 - \dfrac{b^2}{4ac}}\right)$$

The latter can be written as $\,z_1=\sqrt{\dfrac{c}{a}}\big(\cos(\varphi)+ i \sin(\varphi)\big)\,$ where $\,\varphi=\arg(z_1)\,$ and:

$$ \begin{cases} \begin{align} \cos(\varphi) &= -\dfrac{b}{2\sqrt{ac}} \\[5px] \sin(\varphi) &= -\sqrt{1 - \dfrac{b^2}{4ac}} \end{align} \end{cases} $$

Therefore $\,\varphi=2k\pi \pm \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right)\,$, and the $\,\sin(\varphi) \lt 0\,$ condition trims the solution set down to $\,\varphi=2k\pi \color{red}{-} \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right) = (2k-1)\pi + \arccos\left(\dfrac{b}{2\sqrt{ac}}\right)\,$, so in the end:

$$\,\arg\left(\dfrac{z_1}{z_2}\right)= 2 \varphi \;\;\equiv\;\; 2 \arccos\left(\dfrac{b}{2\sqrt{ac}}\right) = 2 \arccos\left(\sqrt{\dfrac{b^2}{4ac}}\right) \pmod{2 \pi} \,$$


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.


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