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...

Here is a solution which is little different, algebraically, though it is "spiritually similar" to  the work of Mario Gallegos:
given that
$z^2 - iz = \vert z - i \vert, \tag{1}$
we first take the modulus of each side, noting that
$\vert z^2 - iz \vert = \vert z(z -i) \vert =  \vert z \vert \vert z -i \vert; \tag{2}$
we thus obtain
$\vert z \vert \vert z -i\vert = \vert \vert z -i \vert \vert = \vert z -i\vert. \tag{3}$
From (3) we see that $\vert z -i \vert \ne 0$ implies $\vert z \vert = 1$; $z =i$ is clearly a solution as may be seen directly from (1); so we look for solutions with $\vert z \vert = 1$, $z \ne i$.  $\vert z \vert = 1$ implies $z = e^{i\theta}$ with $0 \le \theta < 2\pi$; then
$z^2 - iz = e^{2i\theta} - ie^{i\theta} = \cos 2 \theta +i\sin 2\theta -i\cos \theta + \sin \theta.  \tag{4}$
We see from (1) that $z^2 -iz$ must be real; this means the imaginary part of (4) must vanish, or
$\sin 2\theta - \cos \theta = 0; \tag{5}$
since $\sin 2 \theta = 2(\sin \theta)(\cos \theta)$ we have
$2(\sin \theta)(\cos \theta) = \cos \theta: \tag{6}$
if $\cos \theta = 0$, then $\theta = \pi / 2, 3\pi/2 \;$ or $z = \pm i$; we have seen that $z = i$ is a solution and $z = -i$ may be ruled out by direct substitution into (1); thus we may assume $\cos \theta \ne 0$, in which case we have $\sin \theta = 1/2$, or $\Im z = 1/2$.  We now use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to arrive at $\Re z = \cos \theta = \pm \sqrt 3 / 2$, so that in addition to $z =i$, the other two solutions are 
$z = \pm \dfrac{\sqrt 3}{2} + \dfrac{i}{2};  \tag{7}$
all three solution may be verified by direct substitution into (1).
Hope this helps.  Cheerio,
and.as always,
Fiat Lux!!!
