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 $z_1/z_2$ is a number which is

We know that modulus of $e^{i\theta }=1$

$\therefore \frac{Z_1+Z_2}{Z_1-Z_2}=e^{i\theta }=\frac{e^{i\frac{\theta }{2}}}{e^{-i\frac{\theta }{2}}}$

Apply componendo and dividendo.

$\begin{array}{l}\therefore \frac{2z_1}{2z_2}=\frac{e^{i\frac{\theta }{2}}+e^{-i\frac{\theta }{2}}}{e^{\frac{\theta }{z}}-e^{-i\frac{\theta }{2}}}\\ \text{ or }\frac{z_1}{z_2}=\frac{2\cos (\theta /2)}{i.2\sin (\theta /2)}\\ =-i\cot (\theta /2)\end{array}$

This is purely imaginary but when $\theta =k\pi$ then it will be 

zero as cot $(k\pi /2)=0.\text{ i.e. }$ when k is odd.