A point $$z$$ moves in the complex plane such that $$\arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{4}$$, then the minimum value of $$|z - 9\sqrt{2} - 2i|^2$$ is equal to _________.

Let us denote the moving point by $$z=x+iy,\;x,y\in\mathbb R.$$

We are given the condition

$$\arg\!\left(\dfrac{z-2}{z+2}\right)=\dfrac{\pi}{4}.$$

For any complex number $$w=u+iv,$$ the argument equality $$\arg w=\dfrac{\pi}{4}$$ is equivalent to

$$\dfrac{\Im w}{\Re w}=\tan\dfrac{\pi}{4}=1 \quad\text{and}\quad \Re w>0.$$

So we must have

$$\Im\!\left(\dfrac{z-2}{z+2}\right)=\Re\!\left(\dfrac{z-2}{z+2}\right).$$

We now evaluate the real and imaginary parts of $$\dfrac{z-2}{z+2}.$$ Substituting $$z=x+iy$$, we get

$$\frac{z-2}{z+2}=\frac{(x-2)+iy}{(x+2)+iy}.$$

Multiply numerator and denominator by the conjugate of the denominator, $$(x+2)-iy,$$ to rationalise:

$$\frac{(x-2)+iy}{(x+2)+iy}\; \cdot\; \frac{(x+2)-iy}{(x+2)-iy} =\frac{[(x-2)(x+2)+y^{2}] + i[\,y(x+2)-y(x-2)\,]} {(x+2)^2+y^{2}}.$$

Calculating the two parts separately, we find

$$\begin{aligned} (x-2)(x+2)+y^{2}&=x^{2}-4+y^{2},\\ y(x+2)-y(x-2)&=y(x+2-x+2)=4y. \end{aligned}$$

Hence

$$\frac{z-2}{z+2} =\frac{\,x^{2}+y^{2}-4 + i\,4y\,}{x^{2}+y^{2}+4x+4}.$$

Therefore,

$$\Re\!\left(\frac{z-2}{z+2}\right)=\frac{x^{2}+y^{2}-4}{x^{2}+y^{2}+4x+4},\qquad \Im\!\left(\frac{z-2}{z+2}\right)=\frac{4y}{x^{2}+y^{2}+4x+4}.$$

Setting the imaginary part equal to the real part we obtain

$$\frac{4y}{x^{2}+y^{2}+4x+4} =\frac{x^{2}+y^{2}-4}{x^{2}+y^{2}+4x+4}.$$

The common denominator cancels out, leaving

$$4y=x^{2}+y^{2}-4.$$

Rearranging,

$$x^{2}+y^{2}-4y-4=0.$$

To recognise the geometrical locus, we complete the square in $$y$$:

$$y^{2}-4y=(y-2)^{2}-4,$$ so that

$$x^{2}+(y-2)^{2}-4-4=0 \;\Longrightarrow\; x^{2}+(y-2)^{2}=8.$$

Thus the point $$z$$ moves on the circle with centre $$C(0,2)$$ and radius

$$R=\sqrt{8}=2\sqrt2.$$

Next, we must minimise the squared distance

$$|\,z-9\sqrt2-2i\,|^{2}.$$

The fixed point whose distance from $$z$$ is being measured is

$$Q(9\sqrt2,\,2).$$

Notice that both the centre $$C(0,2)$$ of the circle and the point $$Q$$ have the same $$y$$-coordinate $$2$$. Hence the line $$y=2$$ is the line through both the centre and the fixed point. In such a situation, the closest point on the circle to $$Q$$ lies on the segment $$CQ$$ and is simply the point where that line meets the circle on the near side.

The distance between the centre and $$Q$$ is purely horizontal:

$$CQ=9\sqrt2-0=9\sqrt2.$$

For any circle, the minimum distance from an external point to the circle equals the distance from the point to the centre minus the radius. Using

$$\text{Minimum distance}=CQ-R,$$

we obtain

$$\text{Minimum distance}=9\sqrt2-2\sqrt2=7\sqrt2.$$

The question asks for the minimum value of the squared distance, so we square the above:

$$\bigl(7\sqrt2\bigr)^{2}=7^{2}\cdot2=49\cdot2=98.$$

So, the answer is $$98$$.