Question

If $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$ then $\frac{z_1}{z_2}$ is

• A. positive real
• B. negative real
• C. purely imaginary
• D. $0$

Answer 1

The correct option is C purely imaginary

Given that $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$
$\Rightarrow \left|\frac{(\frac{z_1}{z_2}+1)}{(\frac{z_1}{z_2}-1)}\right|=1$
$\Rightarrow \left|\frac{\frac{z_1}{z_2}+1}{\frac{z_1}{z_2}-1}\right|=1$
$[\because \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}]$
$\Rightarrow \frac{|\frac{z_1}{z_2}+1|}{|\frac{z_1}{z_2}-1|}=1$
Let $\frac{z_1}{z_2}=x+iy$
$\frac{|x+iy+1|}{|x+iy-1|}=1$
$\Rightarrow (x+1{)}^2+y^2=(x-1{)}^2+y^2$
$\Rightarrow x^2+1+2x=x^2+1-2x$
$\Rightarrow 4x=0$
$\Rightarrow x=0$
It means $\frac{z_1}{z_2}=iy$
So, clearly $\frac{z_1}{z_2}$ is purely imaginary