If $P$ and $Q$ are represented by the complex numbers $z_{1} a n d z_{2}$ , such that $|\frac{1}{z_{2}} + \frac{1}{z_{1}}| = |\frac{1}{z_{2}} - \frac{1}{z_{1}}|,$ then

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The circumcenter of $Δ O P Q$ is $\frac{1}{2} (z_{1} + z_{2})$

$Δ O P Q$ (where O is the origin) is equilateral

$Δ O P Q$ is right angled

The orthocentre of $Δ O P Q$ is $\frac{1}{3} (z_{1} + z_{2})$

answer is B, C.

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Detailed Solution

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 on both sides, we have
${|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 (z_{1} {\bar{z}}_{2} + {\bar{z}}_{1} z_{2}) = {|z_{1}|}^{2} + {|z_{2}|}^{2} - 2 (z_{1} {\bar{z}}_{2} + {\bar{z}}_{1} z_{2})$

$\Rightarrow 4 (z_{1} {\bar{z}}_{2} + {\bar{z}}_{1} z_{2}) = 0$

$\Rightarrow \frac{z_{1}}{z_{2}} + \frac{{\bar{z}}_{1}}{{\bar{z}}_{2}} = 0$

$\Rightarrow \arg (\frac{z_{1}}{z_{2}}) = \frac{π}{2} = \arg (\frac{z_{1} - 0}{z_{2} - 0})$

The angle between $z_{2}$ , O, and $z_{1}$ is a right angle, taken in order, as shown in the above diagram. Now, the circumcenter of the above diagram will lie on the line $P Q$ as diameter and is represented by C which is the center of $P Q$ , such that $z = (z_{1} + z_{2}) / 2$ , where $z$ is the affix of circumcenter.