Let $z=x+iy$

Since, $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi }{4}\Rightarrow \arg \left(\frac{x+iy-2}{x+iy+2}\right)=\frac{\pi }{4}$

$\Rightarrow \arg \left(\frac{\left(x-2\right)+iy}{\left(x+2\right)+iy}\times \frac{\left(x+2\right)-iy}{\left(x+2\right)-iy}\right)=\frac{\pi }{4}\Rightarrow \arg \left(\frac{\left(x^2-4+y^2\right)+4iy}{{\left(x+2\right)}^2+y^2}\right)=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{4y}{x^2-4+y^2}\right)=\frac{\pi }{4}\Rightarrow 4y=x^2-4+y^2\left[\because \tan \left(\pi /4\right)=1\right]$

$\Rightarrow x^2+y^2-4y+4=8\Rightarrow {\left(x-0\right)}^2+{\left(y-2\right)}^2={\left(2\sqrt{2}\right)}^2,$

Which represents a circle with centre, $A\left(0,2\right)$ and radius, $r=2\sqrt{2}$
Let $P\left(9\sqrt{2},2\right)$ be point which lies outside the circle.

Then, minimum value of ${\left|z-\left(9\sqrt{2}+2i\right)\right|}^2={\left|AP-r\right|}^2$

$={\left(\sqrt{{\left(9\sqrt{2}-0\right)}^2+{\left(2-2\right)}^2}-2\sqrt{2}\right)}^2={\left(7\sqrt{2}\right)}^2=98$