If $z=x+iy, \forall x,y\in R, i^2=-1, xy\ne 0$ and $\left|z\right|=2,$ then the imaginary part of $\frac{z+2}{z-2}$ cannot be

Given, $\left|z\right|=2 \Rightarrow x^2+y^2=4 \& xy\ne 0$

Let, $z=x+iy=2\left(\cos \theta +i\sin \theta \right)$
So, $\frac{z+2}{z-2}=\frac{\left(x+2\right)+iy}{\left(x-2\right)+iy}$

$=\frac{\left(x^2+y^2-4\right)+i\left(y\left(x-2\right)-y\left(x+2\right)\right)}{{\left(x-2\right)}^2+y^2}=\frac{-4y}{8-4x}i$
So, the imaginary part $=-\frac{4y}{8-4x}=\frac{y}{x-2}$

$=\frac{-2\sin \theta }{2-2\cos \theta }=-\cot \frac{\theta }{2},$ $\theta \in \left(0,2\pi \right)$
Now, $\theta \ne \frac{\pi }{2},\mathrm{}\pi ,\frac{3\pi }{2}\Rightarrow \cot \frac{\theta }{2}\ne -1,\mathrm{}0,\mathrm{}1$