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 $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$
$\Rightarrow \left|\frac{(\frac{z_1}{z_2}+1)}{(\frac{z_1}{z_2}-1)}\right|=1\Rightarrow \frac{|\frac{z_1}{z_2}+1|}{|\frac{z_1}{z_2}-1|}=1[\because \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}]$

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
$\frac{z_1}{z_2}$ is purely imaginary