Question

If $z_1,z_2$ are two complex numbers $(z_1\ne z_2)$ 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 $|arg{z}_1-arg{z}_2|=\frac{\pi }{2}$

$|z_1^2-z_2^2|=|{\overline{z}}_1^2+{\overline{z}}_1^2-2{\overline{z}}_1{\overline{z}}_2|$
or $|z_1-z_2||z_1+z_2|=|{\overline{z}}_1-{\overline{z}}_2{|}^2$
or $|z_1+z_2|=|{\overline{z}}_1-{\overline{z}}_2|$
$|z_1+z_2|=|z_1-z_2|$
$\Rightarrow |\frac{z_1}{z_2}+1|=|\frac{z_1}{z_2}-1|$
$\Rightarrow \frac{z_1}{z_2}$ lies on $\perp$ bisector of 1 and -1
$\Rightarrow \frac{z_1}{z_2}$ lies on imaginary axis
$\Rightarrow \frac{z_1}{z_2}$ is purely imaginary
$\Rightarrow arg\left(\frac{z_1}{z_2}\right)=\pm \frac{\pi }{2}$
$|arg\left(z_1\right)-arg\left(z_2\right)|=\frac{\pi }{2}$