Solve the following

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 .