Question

Let $z _1$ and $z _2$ are two complex numbers such that $\left| z _1\right|= a$ and $\left| z _2\right|= b$ satisfying the equation $3 z_1^2-2 z_1 z_2+2 z_2^2=0$ and also $\operatorname{Re}\left(\frac{z_1-2}{z_1+2}\right)=0$. Also $P_1, P_2$ and $O$ are points in complex plane corresponding to $z _1, z _2$ and origin respectively.
Area of triangle $OP _1 P _2$ is

• A. $\sqrt{5}$
• B. $2 \sqrt{5}$
• C. $3 \sqrt{5}$
• D. $4 \sqrt{5}$

Answer 1

Answer: (B)

Also, $ \frac{z_2}{\left|z_2\right|}=\frac{z_1}{\left|z_1\right|} e^{i \theta}$
$\therefore \frac{ z _2}{ z _1}=\frac{\left| z _2\right|}{\left| z _1\right|} e ^{ i \theta} $
$\frac{1 \pm \sqrt{5} i }{2}=\frac{\sqrt{6}}{2} e ^{ i \theta}$
$\therefore \frac{1}{\sqrt{6}} \pm \frac{\sqrt{5}}{\sqrt{6}} i =\cos \theta+ i \sin \theta$
$\cos \theta =\frac{1}{\sqrt{6}} $
$\sin \theta = \pm \frac{\sqrt{5}}{\sqrt{6}} $
$\therefore \Delta OPQ =\left|\frac{1}{2}\right| z _1|| z _2|\sin \theta|=\frac{1}{2}\left| z _1\right| \frac{\sqrt{6}}{2}\left| z _1\right| \frac{\sqrt{5}}{\sqrt{6}}$
Now, $\left|z_1\right|=2$
$\therefore \Delta=\frac{1}{2}(2)\left(\frac{\sqrt{6}}{2}\right)(2)\left(\frac{\sqrt{5}}{\sqrt{6}}\right)=\sqrt{5} $