The area of the polygon, whose vertices are the non-real roots of the equation $$\bar{z} = iz^2$$ is

We need to find the non-real roots of $$\bar{z} = iz^2$$.

Let $$z = x + iy$$, so $$\bar{z} = x - iy$$.

Then $$iz^2 = i(x + iy)^2 = i(x^2 - y^2 + 2ixy) = i(x^2 - y^2) + i^2(2xy) = -2xy + i(x^2 - y^2)$$

Equating real and imaginary parts of $$\bar{z} = iz^2$$:

$$x = -2xy \quad \cdots (1)$$

$$-y = x^2 - y^2 \quad \cdots (2)$$

From equation (1): $$x(1 + 2y) = 0$$

Case 1: $$x = 0$$

From equation (2): $$-y = 0 - y^2 \implies y^2 - y = 0 \implies y(y-1) = 0$$

So $$y = 0$$ giving $$z = 0$$ (real root, excluded), or $$y = 1$$ giving $$z = i$$ (non-real root).

Case 2: $$1 + 2y = 0 \implies y = -\frac{1}{2}$$

From equation (2): $$-\left(-\frac{1}{2}\right) = x^2 - \frac{1}{4}$$

$$\frac{1}{2} = x^2 - \frac{1}{4}$$

$$x^2 = \frac{3}{4}$$

$$x = \pm\frac{\sqrt{3}}{2}$$

This gives two non-real roots: $$z_2 = \frac{\sqrt{3}}{2} - \frac{i}{2}$$ and $$z_3 = -\frac{\sqrt{3}}{2} - \frac{i}{2}$$

The three non-real roots are:

$$z_1 = i = (0, 1)$$

$$z_2 = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

$$z_3 = \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

These form a triangle. We use the formula:

Area $$= \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

$$= \frac{1}{2}\left|0\left(-\frac{1}{2} + \frac{1}{2}\right) + \frac{\sqrt{3}}{2}\left(-\frac{1}{2} - 1\right) + \left(-\frac{\sqrt{3}}{2}\right)\left(1 + \frac{1}{2}\right)\right|$$

$$= \frac{1}{2}\left|0 + \frac{\sqrt{3}}{2}\left(-\frac{3}{2}\right) - \frac{\sqrt{3}}{2}\left(\frac{3}{2}\right)\right|$$

$$= \frac{1}{2}\left|-\frac{3\sqrt{3}}{4} - \frac{3\sqrt{3}}{4}\right|$$

$$= \frac{1}{2} \times \frac{6\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$$

The area of the polygon is $$\dfrac{3\sqrt{3}}{4}$$.

The correct answer is Option B.