Question

If $z_1,z_2$ are the non zero complex root of $z^2-ax+b=0$ such that $|z_1|=|z_2|$, where $a,b$ are complex numbers. If $A\left(z_1\right),B\left(z_2\right)$ and $\angle AOB=\theta$, ${}^{\prime }{O}^{\prime }$ being the origin then $a^2=4b{\cos }^2\frac{\theta }{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

$z^2-ax+b=0$
$z_1,z_{2}aretheroots$
$z_1+z_2=a$
$z_1{z}_2=b.......\left(i\right)$
$\sin ce\theta isanglebetween{z}_{1}and{z}_2$
$Hence$
$z_2=z_1{e}^{i\theta }$
$pluggingin\left(i\right)$
$z_1^2{e}^{i\theta }=b............\left(ii\right)$
$z_1+z_1{e}^{i\theta }=a$
$\Rightarrow z_1(1+e^{i\theta })=a$
$\Rightarrow z_1=\frac{a}{1+e^{i\theta }}$
$from\left(ii\right)$
$\frac{b}{e^{i\theta }}=\frac{a^2}{{(1+e^{i\theta })}^2}$
$=\frac{a^2}{1+e^{2i\theta }+2e^{i\theta }}$
$=\frac{a^2}{1+\cos 2\theta +i\sin 2\theta +2(\cos \theta +i\sin \theta )}$
$=\frac{a^2}{2{\cos }^2\theta +i2\sin \theta \cos \theta +2(\cos \theta +i\sin \theta )}$
$\frac{b}{e^{i\theta }}=\frac{a^2}{2({\cos }^2\theta +i\sin \theta )\cos \theta +2(\cos \theta +i\sin \theta )}$
$\Rightarrow b=\frac{a^2}{2\cos \theta +2}$
$\Rightarrow b=\frac{a^2}{2[1+\cos \theta ]}=\frac{a^2}{2\left[2{\cos }^2\theta /2\right]}=\frac{a^2}{4{\cos }^2\frac{\theta }{2}}$
$\Rightarrow a^2=4b{\cos }^2\frac{\theta }{2}$
$Hencethegivenstatementistrue$