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 $3z_1^2-2z_1{z}_2+2z_2^2=0$ and also $\mathrm{Re}\left(\frac{z_1-2}{z_1+2}\right)=0$. Also $P_1,P_2$ and $O$ are points in complex plane corresponding to $z_1,z_2$ and origin respectively.
Area of triangle $O{P}_1{P}_2$ is

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

Answer 1

Answer: (B)

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