Solve $z^2 - iz = |z - i|$ I have the equation:
$z^2 - iz = |z - i|$
The solutions are $i$, $-\sqrt3/2 + i/2$, $\sqrt3/2 + i/2$
Can someone please walk me through or give me a hint...

I'll try a more conceptual approach.
First thing you should notice is this:
$$
z(z-i) = |z-i|
$$
That should give you the trivial solution $z = i$. Next, recall that $\frac{z}{|z|} = e^{i \arg{z}}$.
$$
\frac{z-i}{|z-i|} = \frac{1}{z} \\
e^{i \arg{(z-i)}} = \frac{1}{z}
$$
It should now be quite clear that $|z| = 1$. Now we just need to figure out the argument/angle. Let $z = e^{i\arg{z}}$.
$$
e^{i \arg{(z-i)}} = e^{-i\arg{z}} \\
\arg{(z-i)} = -\arg{z}
$$
Graphically, what this means is that we want to find all of the angles such that the negative angle on the unit circle is exactly the same as subtracting $i$ (going down one unit) from the original angle. That should give you the other two solutions.
