Modulus of a Complex Number

Given,

$$\left|\frac{z-2}{z+2}\right|=1$$

Hence,

$$|z-2|=|z+2|$$

Let

$$z=x+iy$$

Then,

$$\sqrt{(x-2)^2+y^2}= \sqrt{(x+2)^2+y^2}$$

Squaring both sides,

$$(x-2)^2+y^2=(x+2)^2+y^2$$

$$x^2-4x+4=x^2+4x+4$$

$$x=0$$

Thus,

$$\operatorname{Re}(z)=0$$

Now,

$$|z|^2-4\operatorname {Re}(z)= |z|^2$$

Since

$$z=iy,$$

$$|z|^2=y^2$$

which is not fixed.

Now observe geometrically:

$$|z-2|=|z+2|$$

represents the perpendicular bisector of

$$2$$

and

$$-2,$$

that is,

$$x=0$$

Hence,

$$|z|^2-4\operatorname{Re}(z)=x^2+y^2-4x$$

$$=y^2$$

Taking the least possible value,

$$y=0$$

Thus minimum value is

$$0$$

Therefore, the required answer is

$$\boxed0$$

Let $$z = x + iy$$ where $$x, y \in \mathbb{R}.$$

From $$|z - 1| = 1$$: $$|x - 1 + iy| = 1$$

$$(x-1)^2 + y^2 = 1$$

This is a circle centered at $$(1, 0)$$ with radius 1.

From $$|z - i| = 1$$: $$|x + i(y-1)| = 1$$

$$x^2 + (y-1)^2 = 1$$

This is a circle centered at $$(0, 1)$$ with radius 1.

Finding the intersection of the two circles:

From equation 1: $$x^2 - 2x + 1 + y^2 = 1 \Rightarrow x^2 + y^2 = 2x$$

From equation 2: $$x^2 + y^2 - 2y + 1 = 1 \Rightarrow x^2 + y^2 = 2y$$

Therefore: $$2x = 2y \Rightarrow x = y$$

Substituting into $$x^2 + y^2 = 2x$$:
$$x^2 + x^2 = 2x$$
$$2x^2 = 2x$$
$$x^2 = x$$
$$x(x-1) = 0$$

So $$x = 0$$ or $$x = 1$$

If $$x = 0$$: $$y = 0$$, so $$z = 0$$

Check: $$|0-1| = 1$$  and $$|0-i| = 1$$ 

If $$x = 1$$: $$y = 1$$, so $$z = 1 + i$$

Check: $$|1+i-1| = |i| = 1$$  and $$|1+i-i| = |1| = 1$$ 

For $$z = 0$$: $$|z| = 0$$
For $$z = 1 + i$$: $$|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$$

since  $$\sqrt{2}$$ is only there among 2 solutions, $$\sqrt{2}$$ is the answer

To find the required value, we use the properties of roots of unity, complex rotation, and conjugates step-by-step:

1. Determine the vertices of the equilateral triangle:
The vertices $$z_1, z_2, z_3$$ are inscribed in the circle $$|z| = 2$$ and form an equilateral triangle. Since $$z_1 = 2$$ lies on the positive real axis, the other two vertices are obtained by rotating $$z_1$$ counter-clockwise about the origin by angles of $$\frac{2\pi}{3}$$ and $$\frac{4\pi}{3}$$ respectively.
Using the complex cube root of unity $$\omega = e^{i\frac{2\pi}{3}}$$, we can write:
$$z_1 = 2$$
$$z_2 = 2\omega$$
$$z_3 = 2\omega^2$$

2. Calculate the conjugate of $$z_3$$:
We find the conjugate of $$z_3 = 2\omega^2$$. Since $$\bar{\omega} = \omega^2$$ and $$\overline{\omega^2} = \omega$$, the conjugate simplifies to:
$$\bar{z_3} = \overline{2\omega^2} = 2\omega$$

3. Evaluate the transformed complex number $$z_4$$:
Substitute the values of $$z_1$$, $$z_2$$, and $$\bar{z_3}$$ into the given expression:
$$z_4 = \frac{z_2 \cdot \bar{z_3}}{z_1}$$
$$z_4 = \frac{(2\omega) \cdot (2\omega)}{2}$$
$$z_4 = \frac{4\omega^2}{2} = 2\omega^2$$

Notice that $$z_4$$ is exactly equal to the vertex $$z_3$$.

4. Compute the final absolute squared value:
We need to find the value of $$|z_4 - z_1|^2$$, which is equivalent to finding the squared distance between the vertices $$z_3$$ and $$z_1$$ of our equilateral triangle.
$$|z_4 - z_1|^2 = |2\omega^2 - 2|^2 = 4|\omega^2 - 1|^2$$

We substitute the algebraic value of $$\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$ into the magnitude expression:
$$|\omega^2 - 1| = \left| -\frac{1}{2} - i\frac{\sqrt{3}}{2} - 1 \right| = \left| -\frac{3}{2} - i\frac{\sqrt{3}}{2} \right|$$
$$|\omega^2 - 1|^2 = \left(-\frac{3}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{9}{4} + \frac{3}{4} = \frac{12}{4} = 3$$

Now, multiply this by the scalar factor of 4:
$$|z_4 - z_1|^2 = 4 \times 3 = 12$$

Geometrically, the side length $$s$$ of an equilateral triangle inscribed in a circle of radius $$R = 2$$ is given by $$s = R\sqrt{3} = 2\sqrt{3}$$. The square of the side length (the distance between $$z_3$$ and $$z_1$$) is $$s^2 = (2\sqrt{3})^2 = 12$$.