Compute $\bar z - iz^2 = 0$ $\bar z - iz^2 = 0, i = $ complex unit.
I've found 2 solutions to this, like this:
$x - iy - i(x+iy)^2 = 0$
$i(-x^2+y^2-y)+2xy+x=0$
$2xy + x = 0$ --> $y=-\frac{1}{2}$
$x^2 - y^2 + y = 0$ --> $x_1=\sqrt\frac34$ $x_2=-\sqrt\frac{3}{4}$
Solution 1: $z = \sqrt\frac34 - \frac12i $
Solution 2: $z = -\sqrt\frac34 - \frac12i $
That's great and everything, but Wolfram gives me another solution, which is $z = i$. How do I get that? LINK to wolfram.

$$\bar z=iz^2$$
Equate absolute values:
$$|z|=|\bar z|=|iz^2|=|z|^2\Longrightarrow|z|=0,1.$$
One solution is $z=0$; other solutions satisfy $|z|=1$, so $\bar z=\frac{|z|^2}z=\frac1z$ and the equation simplifies to
$$\frac1z=iz^2,$$
that is,
$$z^3=\frac1i=i^3.$$
One solution obviously is$$z=i.$$ The other cube roots of $i^3$ are $i$ times a cube root of unity, that is,
$$i\omega=i(\cos120^\circ+i\sin120^\circ)=i\left(-\frac12+\frac{\sqrt3}2i\right)=-\frac{\sqrt3}2-\frac12i$$
and
$$i\omega^2=i\bar\omega=i(\cos240^\circ+i\sin240^\circ)=i\left(-\frac12-\frac{\sqrt3}2i\right)=\frac{\sqrt3}2-\frac12i.$$
Alternatively,
$$z^3-i^3=(z-i)(z^2+iz+i^2)=(z-i)(z^2+iz-1)$$
so the last two solutions can be found by solving the quadratic equation
$$z^2+iz-1=0.$$
