Solve equation with a conjugate How to solve this equation:
$$
|z|^2 - 9 - \bar z + 3i = 0
$$
I know that:
$$
|z|^2=z \bar z
$$
$$
|z|=\sqrt{a^2+b^2}
$$
$$
z = a + bi \Rightarrow \bar z = a - bi
$$
But I still don't know to calculate z from that.
 A: Remember that two complex numbers are equal if and only if both their real and imaginary parts are equal. For example, let's take the imaginary part of both sides of that equation, assuming that $z = x + iy$:
$\begin{eqnarray}\Im (|z|^2 - 9 - \bar{z} + 3i) & = & \Im (0 + 0i)\\
\Im (-\bar{z} + 3i) & = & 0\\
\Im (-x+iy+3i) & = & 0\\
y + 3 & = & 0\\
y & = & -3 \end{eqnarray}$
since we know that $|z|$ is always a real number, and $\overline{x + iy} = x - iy$.
Having done that, you can then also look at the real parts:
$\begin{eqnarray}\Re (|z|^2 - 9 - \bar{z} + 3i) & = & \Re (0 + 0i) \\
\Re ((x^2 + y^2) - 9 - (x - iy) + 3i) & = & 0 \\
(x^2 + y^2) - 9 - x & = & 0 \end{eqnarray}$
and since we already know $y$ this is just a quadratic equation in $x$ that should be pretty easy to solve.
A: To solve the equation you must separate the left hand side into its real and imaginary parts and set them both equal to 0.  In this case:
$$
|z|^2-9-Re(z)  = 0\\
Im(z)+3=0
$$
From here it should not be too hard to solve.
A: Be $z=x+iy$ so $|z|^2=x^2+y^2$, $\bar z=x-iy$. Replacing in the equation:$$|z|^2-9-\bar z+3i=0$$ $$x^2+y^2-9-x+iy+3i=0$$
Reagrouping,$$(x^2+y^2-9-x)+i(y+3)=0$$
So $Re(z)=0$ and $Im(z)=0$,$$y+3=0$$$$x^2+y^2-9-x=0$$
Solving for $y$ gives $y=-3$. Replacing in the other equation:
$$x^2+(-3)^2-9-x=0$$
$$x^2-x=0$$
Gives solutions to $x$, $x=0$ $x=1$.
So you have to solutions for z.$z=-3i$ and $z=1-3i$
