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...

Let $z=x+iy$, $z^2-iz=|z-i|$ implies $0=\Im (z^2-iz)=2xy-x$ then $x=0$ or $y=\frac{1}{2}$.
If $x=0$, then we have $-y^2+y=|y-1|$ i.e. $-y(y-1)=|y-1|$ then $y=1$. So $z=i$ satisfies the equation.
Note that $z^2-iz=|z-i|$ implies $|z||z-i|=|z^2-iz|=|z-i|$, then $|z|=1$. So, if $y=\frac{1}{2}$, then $x=\pm\sqrt{1-\left(\frac{1}{2}\right)^2}=\pm\frac{\sqrt{3}}{2}$. If we put $z=\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}$ we get 
\begin{align}
z^2-iz & =z(z-i) \\
& =\left(\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\left(\pm \frac{\sqrt{3}}{2}-i\frac{1}{2}\right) \\
 & =\frac{3}{4}+\frac{1}{4} \\
 & = 1 \\
\end{align}
We can see easily $|z-i|=\left|\pm \frac{\sqrt{3}}{2}-i\frac{1}{2}\right|=1$.
Therefore, $i$, $\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}$ are the roots.
