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...
 A: 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.
A: 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.
A: A simple solution might be to try taking x and y such that z=x+iy and just plug through the algebra.
A: 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!!!
