Solving an equation $z + |z|^2 = 10 - 2i$ How do you solve the equation $z+ |z|^2 = 10 - 2i$ and express the solution in the form $a+bi$?
I feel like this should be easy but I'm having a blank.
 A: Determine the real part and the imaginary part of the equation and assume $z=a+bi$:
$$\operatorname{Re}(z)+|z|^2=10\implies a+a^2+b^2=10$$
$$\operatorname{Im}(z)=-2 \implies b=-2$$
From the first equation, you can determine $a$. Note, that I implicitly used that $|z|^2\in \mathbb{R}$.
A: Write $z=a+ib$ and extract two equations for the real part and imaginary part. Solve for a and b.
Edit: 
$|z|^2+z=10-2i$, we get:
$a^2+b^2+a+ib=10-2i$, hence
$a^2+b^2+a+ib-10+2i=0$
$a^2+b^2+a-10+i(b+2)=0$, therefore the two equations
I. $a^2+b^2+a-10=0$ which is the real part and
II. $b+2=0$ which is the imaginary part.
From II. we get $b=-2$ immediately and can use it in I.
This gives us:
$a^2+a+4-10=0$ 
$a^2+a-6=0$ use a formula for quadratic equations, or see the solutions immediately, since
$a^2+a-6=(a+3)(a-2)=0$ therefore we get two solutions for $a$ namley $a_1=-3$ and $a_2=2$
And we are done.
A: let $$z=x+iy$$ then your equation is
$$x+iy+x^2+y^2=10-2i$$ and you will get the System
$$x+x^2+y^2=10$$
$$y=-2$$
