Question

Let $z_1$ and $z_2$ be complex numbers such that $z_1\ne z_2$ and $\left|z_1\right|=\left|z_2\right|.$ If $z_1$ has positive real part and $z_2$ has negative imaginary part, then show that $\frac{z_1+z_2}{z_1-z_2}$ is purely imaginary.

Answer 1

$z_1=r(\cos \theta +i\sin \theta ),-\frac{\pi }{2}<\theta <\frac{\pi }{2}$
$z_2=r(\cos \varphi +i\sin \varphi ),-\pi <\varphi <0$
$\Rightarrow$ $\frac{z_1+z_2}{z_1-z_2}=-i\cot \left(\frac{0-\varphi }{2}\right),-\frac{\pi }{4}<\frac{\theta -\varphi }{2}<\frac{3\pi }{4}$
Hence purely imaginary