Question

If $P$ and $Q$ are represented by the complex numbers $z_1$ and $z_2$, such that $|\frac{1}{z_2}+\frac{1}{z_1}|=|\frac{1}{z_2}-\frac{1}{z_1}|$, then

• A. $△OPQ$ (where $O$ is the origin) is equilateral
• B. $△OPQ$ is right angled
• C. The circumcentre of $△OPQ$ is $\frac{1}{2}(z_1+z_2)$
• D. The circumcentre of $△OPQ$ is $\frac{1}{3}(z_1+z_2)$

Answer 1

Answer: (B), (C)

The correct options are
B $△OPQ$ is right angled
C The circumcentre of $△OPQ$ is $\frac{1}{2}(z_1+z_2)$

We have,
$|\frac{1}{z_2}+\frac{1}{z_1}|=|\frac{1}{z_2}-\frac{1}{z_1}|\Rightarrow |z_1+z_2|=|z_1-z_2|$

Squaring both sides, we have
$|z_1{|}^2+|z_2{|}^2+2(z_1\overline{z_2}+\overline{z_1}{z}_2)=|z_1{|}^2+|z_2{|}^2-2(z_1\overline{z_2}+\overline{z_1}{z}_2)$
$\Rightarrow 4(z_1\overline{z_2}+\overline{z_1}{z}_2)=0$
$\Rightarrow \frac{z_1}{z_2}+\frac{\overline{z_1}}{z_2}=0$
$\Rightarrow \arg \left(\frac{z_1}{z_2}\right)=\pm \frac{\pi }{2}=\arg \left(\frac{z_1-0}{z_2-0}\right)$

$\therefore △POQ$ is a rightangled triangle
$\Rightarrow$ circumcentre $=\frac{1}{2}(z_1+z_2)$