Question:
If $\frac{z-1}{z+1}$ is purely imaginary number $(z \neq-1)$, find the value of $|z|$.
Solution:
Let $z=x+i y$.
Then,
$\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$
$=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$
$=\frac{x^{2}+x-i x y-x-1+i y+i x y+i y-i^{2} y^{2}}{(x+1)^{2}-i^{2} y^{2}}$
$=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+1+2 x+y^{2}} \quad\left[\because i^{2}=-1\right]$
If $\frac{z-1}{z+1}$ is purely imaginary number, then
$\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0$
$\Rightarrow x^{2}+y^{2}-1=0$
$\Rightarrow x^{2}+y^{2}=1$
$\Rightarrow|z|^{2}=1$
$\Rightarrow|z|=1$
Thus, the value of $|z|$ is 1 .