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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.

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  • $\begingroup$ What is raised at the power $1/2$ exactly? This is not clear. $\endgroup$ – Bernard Sep 3 '17 at 10:54
  • $\begingroup$ It is 2Cos inverse[(b^2/4ac)^1/2] $\endgroup$ – Samar Imam Zaidi Sep 3 '17 at 10:58
  • $\begingroup$ OK. Note the function arccos is coded \arccos just like in LaTeX. $\endgroup$ – Bernard Sep 3 '17 at 11:06
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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?

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  • $\begingroup$ I missed the concept that roots are complex conjugate, after that it becomes easy $\endgroup$ – Samar Imam Zaidi Sep 3 '17 at 11:05
  • $\begingroup$ @Samar Imam Zaidi Well done! $\endgroup$ – Robert Z Sep 3 '17 at 11:10
  • $\begingroup$ Please make the following Modification $$z_2=-\frac{b}{2a}+i\frac{\sqrt{4ac-b^2}}{2a}=\overline{z_1}.$$ As $\ z_2$ belongs to second quadrant, rest is Okay $\endgroup$ – Samar Imam Zaidi Sep 3 '17 at 13:15
  • $\begingroup$ @Samar Imam Zaidi Done. Thanks. $\endgroup$ – Robert Z Sep 3 '17 at 13:18

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