Question

If $z_1,z_2$ are two different complex numbers satisfying $|z_1^2-z_2^2|=|{\overline{z}}_1^2+{\overline{z}}_2^2-2{\overline{z}}_1{\overline{z}}_2|,$ then

• A. $\frac{z_1}{z_2}$ is purely imaginary.
• B. $\frac{z_1}{z_2}$ is purely real.
• C. $|\arg z_1-\arg z_2|=\pi$
• D. $|\arg z_1-\arg z_2|=\frac{\pi }{2}$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{z_1}{z_2}$ is purely imaginary.
D $|\arg z_1-\arg z_2|=\frac{\pi }{2}$

Given : $|z_1^2-z_2^2|=|{\overline{z}}_1^2+{\overline{z}}_2^2-2{\overline{z}}_1{\overline{z}}_2|$
$\Rightarrow |z_1-z_2||z_1+z_2|=|{\overline{z}}_1-{\overline{z}}_2{|}^2\Rightarrow |z_1-z_2||z_1+z_2|=|{\overline{z}}_1-{\overline{z}}_2||z_1-z_2|\Rightarrow |z_1+z_2|=|{\overline{z}}_1-{\overline{z}}_2|\Rightarrow |z_1+z_2|=|z_1-z_2|\Rightarrow |\frac{z_1}{z_2}+1|=|\frac{z_1}{z_2}-1|$

$\frac{z_1}{z_2}$ lies on the perpendicular bisector of the line segment joining the points $(1,0)$ and $(-1,0)$
So,
$Re\left(\frac{z_1}{z_2}\right)=0$
Therefore,
$\arg \left(\frac{z_1}{z_2}\right)=\pm \frac{\pi }{2}\Rightarrow |\arg \left(z_1\right)-\arg \left(z_2\right)|=\frac{\pi }{2}$