1. Given equation

We need the non-real roots of $\overline z = i z^2.$

Let $z = x+iy,$ so that $\overline z = x-iy.$

Also, $z^2 = (x+iy)^2 = x^2-y^2+2ixy.$ Hence, $iz^2 = i(x^2-y^2+2ixy)= i(x^2-y^2)-2xy.$ So $iz^2 = -2xy + i(x^2-y^2).$

Now equate real and imaginary parts in $x-iy = -2xy + i(x^2-y^2).$

This gives: $x = -2xy \qquad ...(1)$ $-y = x^2-y^2 \qquad ...(2)$

2. Solve the system

From (1): $x=-2xy \implies x(1+2y)=0.$ So either

• Case 1: $x=0$
• Case 2: $1+2y=0 \implies y=-\frac12$

3. Case 1: $x=0$

Using (2): $-y = 0-y^2 = -y^2$ $y = y^2$ $y(y-1)=0.$ So $y=0 \text{ or } y=1.$ Thus roots are $z=0,\quad z=i.$ Among these, the non-real one is $i$.

4. Case 2: $y=-\frac12$

Substitute into (2): $-\left(-\frac12\right)=x^2-\left(\frac14\right)$ $\frac12 = x^2-\frac14$ $x^2 = \frac34$ $x=\pm \frac{\sqrt3}{2}.$ So the roots are $z=\frac{\sqrt3}{2}-\frac{i}{2}, \qquad z=-\frac{\sqrt3}{2}-\frac{i}{2}.$ These are also non-real.

Hence the non-real roots are: $i,\quad \frac{\sqrt3}{2}-\frac{i}{2},\quad -\frac{\sqrt3}{2}-\frac{i}{2}.$

These are three points in the Argand plane: $A=(0,1),\quad B=\left(\frac{\sqrt3}{2},-\frac12\right),\quad C=\left(-\frac{\sqrt3}{2},-\frac12\right).$

5. Find the area of the polygon

These three points form a triangle.

Observe that $BC$ is horizontal with $\text{base } BC = \frac{\sqrt3}{2}-\left(-\frac{\sqrt3}{2}\right)=\sqrt3.$

Its y-coordinate is $-\frac12$, while point $A$ has y-coordinate $1$. So the height from $A$ to line $BC$ is $1-\left(-\frac12\right)=\frac32.$

Therefore area is $\text{Area} = \frac12 \times \sqrt3 \times \frac32 = \frac{3\sqrt3}{4}.$

6. Check options

$\frac{3\sqrt3}{4}$ corresponds to Option A.

7. Comparison with stored correct answer

Stored correct answer is A, which matches our result.