Question

If $\frac{z-1}{z+1}$ is purely imaginary
number $(z\ne -1)$,find the value of $|z|$

Answer 1

If z is puerly imaginary number then $\overline{}=-z$
$\therefore \overline{\left(\frac{z-1}{z+1}\right)}=-\frac{z-1}{z+1}$
$\Rightarrow \frac{\overline{z-1}}{\overline{z+1}}=-\frac{z-1}{z+1}$
$\Rightarrow \left(\frac{\overline{z}-1}{\overline{z}+1}\right)=-\frac{z-1}{z+1}$
$\Rightarrow (\overline{z}-1)(z+1)=-(z-1)(\overline{z}+1)$
$\Rightarrow (\overline{zz}+\overline{z}-z-1=-\overline{zz}-z+\overline{z}+1$
$\Rightarrow 2z\overline{z}=2$
$\Rightarrow 2{\parallel z\parallel }^2=2$
$\Rightarrow |z|=\pm 1$
$\therefore |z|=1$