If $\frac{z-2}{z+2}\left(z\ne -2\right)$ is purely imaginary then $|z|$ is equal to

Let $z=x+iy$ 

Then, $\frac{z-2}{z+2}=\frac{x+iy-2}{x+iy+2}=\frac{\left(x-2\right)+iy}{\left(x+2\right)+iy}$

$=\frac{\left[\left(x-2\right)+iy\right]\left[\left(x+2\right)+iy\right]}{{\left(x+2\right)}^2+y^2}$

$=\frac{\left(x^2+y^2-4\right)+i\left(4y\right)}{{\left(x+2\right)}^2+y^2}$

Since $\frac{z-2}{z+2}$ is purely imaginary,

$\therefore x^2+y^2-4=0$

$\Rightarrow x^2+y^2=4\Rightarrow {\left|z\right|}^2=4\Rightarrow \left|z\right|=2.$