Question

If the number $\frac{z-1}{z+1}$ is a pure imaginary, then prove that $\left|z\right|=1$.

Answer 1

Let $\frac{z-1}{z+1}$ be purely imaginary. Then

$\frac{x^2+y^2-1}{{(x+1)}^2+y^2}=0$,

Whence $x^2+y^2-1=0$, that is

$\left|z\right|=\sqrt{x^2+y^2}=1$. This completes the proof.