Question

If $z_1,z_2$ are 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, find $arg\left(\frac{z_1}{z_2}\right)=$

• A. ${\cos }^{-1}{\left(\frac{b^2}{2ac}\right)}^{\frac{1}{2}}$
• B. $2{\cos }^{-1}{\left(\frac{b^2}{4ac}\right)}^{\frac{1}{2}}$
• C. ${\cos }^{-1}{\left(\frac{b^2}{4ac}\right)}^{\frac{1}{2}}$
• D. $2{\cos }^{-1}{\left(\frac{b^2}{2ac}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is B $2{\cos }^{-1}{\left(\frac{b^2}{4ac}\right)}^{\frac{1}{2}}$

$\frac{z_1}{z_2}=e^{i\theta }$

$\frac{z_1+z_2}{z_1-z_2}=\frac{e^{i\theta }+1}{e^{i\theta }-1}=\frac{\cos \theta +1+i\sin \theta }{\cos \theta -1+i\sin \theta }$

$=\frac{2\cos \frac{\theta }{2}{e}^{i\frac{\theta }{2}}}{-2{\sin }^2\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$

$\frac{z_1+z_2}{z_1-z_2}=\frac{\cot \frac{\theta }{2}}{i}$

$i\tan \frac{\theta }{2}=\left(\frac{z_1-z_2}{z_1+z_2}\right)$
$-{\tan }^2\frac{\theta }{2}={\left(\frac{z_1-z_2}{z_1+z_2}\right)}^2$

$\Rightarrow 1-se{c}^2\frac{\theta }{2}=\frac{\frac{b^2}{a^2}-\frac{4c}{a}}{\frac{b^2}{a^2}}$

$\Rightarrow 1-{\sec }^2\frac{\theta }{2}=1-\frac{4ac}{b^2}$

$\Rightarrow {\cos }^2\frac{\theta }{2}=\frac{b^2}{4ac}$

$\Rightarrow \cos \frac{\theta }{2}=\sqrt{\frac{b^2}{4ac}}$

$\Rightarrow \theta =2{\cos }^{-1}\sqrt{\frac{b^2}{4ac}}$