Euler's Formula

Let $$\omega = e^{i \frac{2\pi}{3}}$$ denote the principal cube root of unity. Since the vertices $$z_1, z_2, z_3$$ form a counter-clockwise equilateral triangle inscribed in a circle centered at the origin, they are symmetrically distributed. This geometric configuration allows us to express $$z_2$$ and $$z_3$$ in terms of $$z_1$$ using the standard rotation operator about the origin:
$$z_2 = z_1 e^{i \frac{2\pi}{3}} = \omega z_1$$
$$z_3 = z_1 e^{i \frac{4\pi}{3}} = \omega^2 z_1$$

We now analyze the algebraic descriptions of the transformed vertices $$w_1, w_2, w_3$$ using the conditions specified in the problem statement. The first vertex $$w_1$$ is obtained by rotating $$z_1$$ counter-clockwise by an angle of $$\frac{\pi}{3}$$ about the origin, which corresponds to multiplication by the rotation operator $$e^{i \frac{\pi}{3}}$$:
$$w_1 = z_1 e^{i \frac{\pi}{3}}$$

The second vertex $$w_2$$ is scaled radially by a factor of 2 while keeping its argument fixed, which means its magnitude doubles:
$$w_2 = 2 z_2 = 2 \omega z_1 = 2 z_1 e^{i \frac{2\pi}{3}}$$

The third vertex remains stationary, meaning its position vector is unaltered:
$$w_3 = z_3 = \omega^2 z_1 = z_1 e^{i \frac{4\pi}{3}}$$

The center of mass $$z_0$$ of the new triangle is the arithmetic mean of its three vertices. Substituting our expressions for $$w_1, w_2, w_3$$ into the centroid formula yields:
$$z_0 = \frac{w_1 + w_2 + w_3}{3} = \frac{z_1}{3} \left( e^{i \frac{\pi}{3}} + 2 e^{i \frac{2\pi}{3}} + e^{i \frac{4\pi}{3}} \right)$$

We simplify the expression inside the parentheses by converting the exponential forms into rectangular coordinates using Euler's formula:
$$e^{i \frac{\pi}{3}} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$$
$$2 e^{i \frac{2\pi}{3}} = 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -1 + i \sqrt{3}$$
$$e^{i \frac{4\pi}{3}} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$$

Summing these rectangular terms together simplifies the centroid expression significantly:
$$e^{i \frac{\pi}{3}} + 2 e^{i \frac{2\pi}{3}} + e^{i \frac{4\pi}{3}} = \left( \frac{1}{2} - 1 - \frac{1}{2} \right) + i \left( \frac{\sqrt{3}}{2} + \sqrt{3} - \frac{\sqrt{3}}{2} \right) = -1 + i \sqrt{3}$$

Substituting this back into the formula for $$z_0$$ and taking the absolute value on both sides gives:
$$z_0 = \frac{z_1}{3} (-1 + i \sqrt{3}) \implies \vert z_0 \vert = \frac{\vert z_1 \vert}{3} \vert -1 + i \sqrt{3} \vert$$

Since the original circle has a radius of 2, the magnitude of vertex $$z_1$$ is given by $$\vert z_1 \vert = 2$$. Evaluating the magnitude of the scalar factor yields $$\vert -1 + i \sqrt{3} \vert = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$$. This gives :
$$\vert z_0 \vert = \frac{2}{3} \cdot 2 = \frac{4}{3}$$

To evaluate the expression $$\vert z_1 + 2z_2 + 3z_3 \vert^2$$, we substitute the rotation relationships $$z_2 = \omega z_1$$ and $$z_3 = \omega^2 z_1$$ directly into the vector sum:
$$z_1 + 2z_2 + 3z_3 = z_1 (1 + 2\omega + 3\omega^2)$$

Using the fundamental identity for the cube roots of unity, $$1 + \omega + \omega^2 = 0$$, we can rewrite and reduce the polynomial expression inside the parentheses:
$$1 + 2\omega + 3\omega^2 = (1 + \omega + \omega^2) + (\omega + 2\omega^2) = 0 + \omega + 2\omega^2 = \omega + 2\omega^2$$

Applying the substitution $$\omega^2 = -1 - \omega$$ simplifies the expression further:
$$\omega + 2(-1 - \omega) = \omega - 2 - 2\omega = -2 - \omega$$

We now calculate the squared magnitude of this simplified expression. Substituting the standard rectangular form $$\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$$ yields:
$$-2 - \omega = -2 - \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -\frac{3}{2} - i \frac{\sqrt{3}}{2}$$

The squared magnitude of this complex factor is computed by adding the squares of its real and imaginary parts:
$$\vert -2 - \omega \vert^2 = \left( -\frac{3}{2} \right)^2 + \left( -\frac{\sqrt{3}}{2} \right)^2 = \frac{9}{4} + \frac{3}{4} = \frac{12}{4} = 3$$

Finally, we find the value of the complete target expression by combining the independent components:
$$\vert z_0 \vert \cdot \vert z_1 + 2z_2 + 3z_3 \vert^2 = \vert z_0 \vert \cdot \vert z_1 \vert^2 \cdot \vert 1 + 2\omega + 3\omega^2 \vert^2 = \frac{4}{3} \cdot (2)^2 \cdot 3 = 4 \cdot 4 = 16$$