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.

Did's suggestion is excellent:  setting $z = re^{i\theta}$ we have $\bar z = re^{-i\theta}$ and $z^2 = r^2e^{2i\theta}$  whence
$re^{-i\theta} = i r^2 e^{2i\theta}; \tag{1}$
using
$i = e^{i \pi/2} \tag{2}$
yields
$r e^{-i\theta} = r^2 e^{i(2\theta + \pi/2)}; \tag{3}$
taking moduli,
$r = r^2 \Rightarrow r = 0, 1; \tag{4}$
when $r = 0$, $z = 0$; for $r = 1$,
$e^{-\theta} = e^{2\theta+ \pi/2}͵ \tag{5}$
whence
$e^{i(3\theta + \pi/2)} = e^{\pi i/2}(e^{i\theta})^3 = i(e^{i\theta})^3 = 1, \tag{6}$
or
$(e^{i\theta})^3 = -i. \tag{7}$
When $r = 1$, $z = re^{i\theta} = e^{i\theta}$, so (7) tells us that $z$ must be a cube root of $-i$; furthermore, we can work backwards from (7): if
$z^3 = -i, \tag{8}$
then, taking moduli,
$\vert z \vert^3 = 1, \tag{9}$
so that
$\vert z \vert = 1; \tag{10}$
then (8) implies
$i z^2 = i \vert z \vert^2 z^2 = i \bar z z z ^2 = i \bar z z^3 = i (-i) \bar z = \bar z; \tag{11}$
we see the non-zero solutions of  
$\bar z = i z^2. \tag{12}$
are precisely the three cube roots of $-i = e^{3\pi i/2}$, viz. $e^{3 \pi i/6} = e^{\pi i / 2} = e^{-3 \pi i / 2} = i$, $e^{(1/2 + 2/3) \pi i} = e^{7 \pi i / 6} = e^{-5 \pi i / 6}$, $e^{(1/2 + 4/3) \pi i} = e^{11 \pi i / 6} = e^{- \pi i / 6}$.
We check this result for $z = e^{7 \pi i / 6}$:
$z = e^{7 \pi i / 6} = -\dfrac{\sqrt{3}}{2} - \dfrac{1}{2} i \tag{13}$
$z^2 = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i; \tag{14}$
$iz^2 = -\dfrac{\sqrt{3}}{2} +  \dfrac{1}{2} i = \bar z;  \tag{15}$
a similar calculation validates $z = e^{11 \pi i / 6}$.
It is easy to see that $i(i)^2 = -i$ and the solution $z = 0$ "checks itself", as it were.  The complete solution set is thus
$\{ 0, i,  e^{7 \pi i / 6},  e^{11 \pi i / 6} \}. \tag{16}$
Note: Not to put too fine a point on it, but for the sake of accuracy it should be observed that $2$ of heropup's proposed solutions, namely $e^{2 \pi i / 3}$ and $e^{-2 \pi i / 3} = e^{4 \pi i / 3}$, do not satisfy the given equation (12):
$e^{2 \pi i / 3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i; \tag{17}$
$(e^{2 \pi i / 3})^2 = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i; \tag{18}$
$\overline{e^{2 \pi i / 3}} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i \ne \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} i = i (e^{2 \pi i / 3})^2, \tag{19}$
with a similar calculation for $e^{4 \pi i / 3}$.  One can also simply write
$(e^{2 \pi / 3})^3 = e^{2 \pi i} = 1 \ne -i, \tag{20}$
etc.  I do not know exactly where heropup's error lies, but we can see from the above his proposed solutions do not all satisfy $\bar z = i z^2$.  End of Note.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
