Let $A$ be a complex number and $B$ be a real number. Prove that $\mid z^2\mid+Re(Az)+B=0$ can only have a solution iff $\mid A^2 \mid \ge 4B$. Been stumped on this question for a while. I tried letting $z=\mid z \mid \cdot e^{i \alpha}$ and $A=\mid A \mid \cdot e^{i\beta}$ -- assuming that $\alpha$ and $\beta$ were the arguments of $z$ and $A$ respectively. 
Substituting gave me
$\mid z\mid^2+\mid z\mid \mid A \mid cos(\alpha+\beta)+B=0$.
I knew, since $\mid z\mid$ was real, the discriminant had to be greater than or equal to $0$. 
$(\mid A\mid cos(\alpha+\beta))^2-4B > 0$
$\mid A^2 \mid cos^2(\alpha+\beta) \gt 4B$
That's as close as I could get. Is there something I'm overlooking? 
 A: Continuing your answer:
$$
{(\cos{(\alpha +\beta)})}^{2}\leq 1\\
4B\leq |A^2|{(\cos{(\alpha+\beta)})}^{2}\leq |A^2| \\
|A^2|\geq 4B
$$
A: (1) : If $|z|^2+Re(Az) =-B$ then $$0 \leq |2z+A|^2$$ $$=4|z|^2 +4Re(Az)+A^2$$ $$=-4B+|A|^2.$$(2)  :  If $|A|^2-4B \geq 0 $, let $z=(\sqrt { |A|^2-4B}  -A)/2$. Then $|z|^2+Re(Az)+B=0.$
A: Note that $A=|A|e^{i\alpha}$ and $B\in {\mathbb R}$ are given, and we want to find a $z=re^{i\phi}$ such that your equation holds. In terms of the new variables $(r,\phi)$ this means that
$$r^2+|A|r\cos(\alpha+\phi)+B=0\tag{1}$$
should have at least one solution $(r,\phi)$ with $r\geq0$, $\phi\in{\mathbb R}$. The discriminant is
$$D=|A|^2\cos^2(\alpha+\phi)-4B\ .$$
If $|A|^2<4B$ then $D$ is negative, and we have no solution. If $|A|^2\geq 4B$ choose $\phi:=\pi-\alpha$ and obtain $D=|A|^2-4B\geq0$. Furthermore we then get
$$r={1\over2}\bigl(|A|\pm\sqrt{D}\bigr)\ ,$$
and at least one of these two values is $\geq0$.
Update: In the case $|A|^2\geq4B$ putting $\phi:=\pi-\alpha$ always leads to an admissible $z$, but there may be other good $\phi$'s, "depending on available space". The question was about the existence of solutions, and did not ask for producing all of them.
