Question

If $z_1$ and $z_2$ are two complex numbers satisfying the equation $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$, then $\frac{z_1}{z_2}$ is a number which is

• A. Purely real
• B. Of unit modulus
• C. Purely imaginary
• D. None of these

Answer 1

Answer: (C)

Purely imaginary

$\begin{array}{c}\left|\frac{z_1+z_2}{z_1-z_2}\right|=1\\ now,\\ \Rightarrow \left|z_1+z_2\right|=\left|z_1-z_2\right|\\ \Rightarrow {\left|z_1+z_2\right|}^2={\left|z_1-z_2\right|}^2\\ \Rightarrow \left(z_1+z_2\right)\left(z_1+z_2\right)=\left(z_1-z_2\right)\left(z_1-z_2\right)\\ \Rightarrow {\left|z_1\right|}^2+{\left|z_2\right|}^2+z_1{z}_2+z_2{z}_1={\left|z_1\right|}^2+{\left|z_2\right|}^2-z_1{z}_2-z_2{z}_1\\ \Rightarrow 2\left(z_1{z}_2+z_2{z}_1\right)=0\\ \Rightarrow z_1{z}_2+z_2{z}_1=0\\ \Rightarrow \frac{z_1}{z_2}+\frac{z_1}{z_2}=0\\ \Rightarrow 2Re\left(\frac{z_1}{z_2}\right)=0\\ \Rightarrow Re\left(\frac{z_1}{z_2}\right)=\frac{z_1}{z_2}\\ =\frac{z_1}{z_2}\end{array}$

hence, $\frac{z_1}{z_2}$ is purely imazinary.

so, option $C$ is correct answer.