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)

$\frac{z_1}{z_2}$ is purely imaginary
$|arg{z}_1-arg{z}_2|=\frac{\pi }{2}$

$|{z_1}^2-{z_2}^2|=|{{\overline{z}}_1}^2+{{\overline{z}}_2}^2-2{\overline{z}}_1{\overline{z}}_2|⟹|{z_1}^2-{z_2}^2|=\left|{(\overline{z_1}-\overline{z_2})}^2\right|=\left|{(z_1-z_2)}^2\right|⟹|z_1-z_2||z_1+z_2|=|z_1-z_2||z_1-z_2|\because z_1\ne z_2\therefore \left|z_1{+z}_2\right|=\left|z_1{-z}_2\right|⟹|\frac{z_1}{z_2}+1|=|\frac{z_1}{z_2}-1|⟹Re\left(\frac{z_1}{z_2}\right)=0\therefore \frac{z_1}{z_2}ispurelyimaginary.\&\left|arg\frac{z_1}{z_2}\right|=\frac{\Pi }{2}$
Ans: A & D