If $\text{'}\mathrm{Z}\text{'}$ lies on the circle $\left|\mathrm{Z}-2\mathrm{i}\right|=2\sqrt{2}$ then the value of $\mathrm{Arg}\left(\frac{\mathrm{Z}-2}{\mathrm{Z}+2}\right)$ is equal to

$\Rightarrow$Z lies on $|\mathrm{z}-2\mathrm{i}|=2\sqrt{2}$

$\Rightarrow \mathrm{Z}$ lies on a circle with centre $\left(0, 2\right)$  having radius $2\sqrt{2}$

 The circle pastes through  $\mathrm{A}(-2,0),\mathrm{B}(2,0)$ slope of  $\mathrm{CA}=\frac{2-0}{0+2}=1$ slope of  $\mathrm{CB}=\frac{2-0}{0-2}=-1$

$\Rightarrow \mathrm{ }CA\perp CB\Rightarrow \mathrm{\angle }ACB={90}^{\circ }$

The chord AB subtends ${90}^0$ at centre 

$\Rightarrow$ Angle subtended by chord  AB on the circumference $= {45}^0$

$$\begin{gathered} \angle \mathrm{APB} = {45}^0 \\ \Rightarrow \mathrm{Arg} \left(\frac{\mathrm{Z} - 2}{\mathrm{Z} + 2}\right) = \frac{\mathrm{\pi }}{4} \end{gathered}$$