The area of the polygon, whose vertices are the non-real roots of the equation z=iz^2, is:

Question

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

Accepted Answer

Let $z=x+iy$. Then $x-iy=i(x^2-y^2+2xyi)=-2xy+i(x^2-y^2)$, giving:
$$x=-2xy \quad\text{and}\quad -y=x^2-y^2.$$

$x=-2xy \implies x=0$ or $y=-\dfrac12$.

$x=0$: $-y=-y^2 \implies y=0,1$. Points: $0$ (real, excluded) and $i$.

$y=-\dfrac12$: $\dfrac12=x^2-\dfrac14 \implies x^2=\dfrac34 \implies x=\pm\dfrac{\sqrt3}2$. Points: $\dfrac{\sqrt3}2-\dfrac i2$ and $-\dfrac{\sqrt3}2-\dfrac i2$.

The three non-real roots are $i,\ \dfrac{\sqrt3}2-\dfrac i2,\ -\dfrac{\sqrt3}2-\dfrac i2$, forming a triangle. Using the shoelace formula:
$$\text{Area}=\frac12\left|0(-\tfrac12-(-\tfrac12))+\tfrac{\sqrt3}2(-\tfrac12-1)+\left(-\tfrac{\sqrt3}2\right)(1-(-\tfrac12))\right|=\frac{3\sqrt3}4.$$

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