Distance Between Two Complex Numbers

$$\text{Re}\left(\frac{z_1 + z_2}{z_3}\right) = -1$$
Let $$z_1 + z_2 = z_s$$ (the resultant vector of $$z_1$$ and $$z_2$$). The equation becomes:
$$\text{Re}\left(\frac{z_s}{z_3}\right) = -1 \implies \frac{|z_s|}{|z_3|} \cos(\theta) = -1$$
Where $\theta$ is the angle between the vector $$(z_1 + z_2)$$ and $$z_3$$. Since $$|z_3| = 2$$:

$$|z_1 + z_2| \cos(\theta) = -2$$

For any triangle inscribed in a circle, the product of its sides $$|z_1 - z_2| \cdot |z_2 - z_3| \cdot |z_3 - z_1|$$ is maximized when the triangle is equilateral.

In an equilateral triangle, the angle between any two position vectors from the origin is $$120^\circ$$.
The resultant $$|z_1 + z_2|$$ of two vectors of length 2 at $$120^\circ$$ is exactly $$2$$
The vector $$(z_1 + z_2)$$ points directly opposite to the third vertex $$z_3$$, meaning the angle $$\theta$$ between them is $$180^\circ$$.

$$|z_1 + z_2| \cos(\theta) = 2 \cdot \cos(180^\circ) = 2(-1) = -2$$
For an equilateral triangle inscribed in a circle of radius R = 2, the side length s is:
$$s = R\sqrt{3} = 2\sqrt{3}$$The required product is simply $$s^3$$
$$\text{Maximum Value} = (2\sqrt{3})^3 = 8 \cdot 3\sqrt{3} = 24\sqrt{3}$$

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 $$s^2 = (2\sqrt{3})^2 = 12$$.

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

$$\frac{z - 5}{z + 5} = \frac{(x - 5) + iy}{(x + 5) + iy} \cdot \frac{(x + 5) - iy}{(x + 5) - iy} = \frac{(x^2 + y^2 - 25) + i(10y)}{(x + 5)^2 + y^2}$$

Given $$\text{Re}\left(\frac{z - 5}{z + 5}\right) = 0$$:

$$\frac{x^2 + y^2 - 25}{(x + 5)^2 + y^2} = 0 \implies x^2 + y^2 = 25 \implies \vert z \vert = 5$$

This defines a circle centered at the origin with radius $$R = 5$$. Let the given fixed point be $$A = 3 + 4i$$. We check its position relative to the circle:

$$\vert A \vert = \sqrt{3^2 + 4^2} = 5$$

Since $$\vert A \vert = 5$$, the point $$A$$ lies directly on the circumference of the circle.

By circle geometry, the maximum distance between any two points on a circle occurs when they are diametrically opposite. Therefore, the point $$z_m$$ that maximizes the distance $$\vert z - A \vert$$ must be the opposite endpoint of the diameter passing through $$A$$.

$$z_m = -A = -(3 + 4i) = -3 - 4i$$

Extracting the real and imaginary parts of $$z_m$$:

$$\text{Re}(z_m) = -3, \quad \text{Im}(z_m) = -4$$

$$\text{Re}(z_m) \cdot \text{Im}(z_m) = (-3) \cdot (-4) = 12$$