Question

If $\frac{z-1}{z+1}$ is purely Imaginary, then prove that $\left|z\right|=1$

Answer 1

Re $\left(\frac{z-1}{z+1}\right)=0$
$\Rightarrow$ $\frac{z-1}{z+1}+\left(\frac{\overline{z-1}}{z+1}\right)=0$
$\Rightarrow$ $\frac{z-1}{z+1}+\frac{\overline{z}-1}{\overline{z}+1}=0$
$\Rightarrow$ $z\overline{z}-\overline{z}+z-1+z\overline{z}-z+\overline{z}-1=0$
$\Rightarrow$ $z\overline{z}=1$
$\Rightarrow$ ${\left|z\right|}^2=1$
$\Rightarrow$ $\left|z\right|=1$