Question

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

Answer 1

Given : $\frac{z-1}{z+1}$ is purely imaginary number, $(z\ne -1)$

If $\alpha$ is purely imaginary number
$\alpha +\overline{\alpha }=0$
So,
$\left(\frac{z-1}{z+1}\right)+\overline{\left(\frac{z-1}{z+1}\right)}=0$
$\Rightarrow \left(\frac{z-1}{z+1}\right)+\left(\frac{\overline{z}-1}{\overline{z}+1}\right)=0$
$\Rightarrow \frac{(z-1)(\overline{z}+1)+(z+1)(\overline{z}-1)}{(z+1)(\overline{z}+1)}=0$
$\Rightarrow \frac{z\overline{z}-\overline{z}+z-1+z\overline{z}+\overline{z}-z-1}{(z+1)(\overline{z}+1)}=0$
$\Rightarrow \frac{2z\overline{z}-2}{(z+1)(\overline{z}+1)}=0$
$\Rightarrow 2z\overline{z}-2=0[\because z\ne -1\Rightarrow \overline{z}\ne -1]$
$\Rightarrow z\overline{z}=1$
$\Rightarrow |z{|}^2=1[\because z\overline{z}=|z{|}^2]$
$\therefore |z|=1[\because |z|\ge 0]$

Hence, the value of $|z|=1$