Solve $|z^2|+\text{Re}(az)+b = 0.$ 
Fix $a\in\mathbb{C}$ and $b\in\mathbb{R}.$ Show that the equation $|z^2|+\text{Re}(az)+b = 0$ has a solution iff $|a^2|\geq 4b.$ When solutions exist, show the solution set is a circle.

A seemingly easy problem. Letting $z=x+yi$ and $a = c+di$ the equation can be rewritten as $$x^2+y^2+cx-dy+b = 0.$$
I don't know how to proceed. 

It is a matter of completing squares:
\begin{align*}
& x^{2} + y^{2} + cx - dy + b = 0\Longleftrightarrow \left(x^{2} + cx + \frac{c^{2}}{4}\right) + \left(y^{2} - dy + \frac{d^{2}}{4}\right) + b - \frac{c^{2} + d^{2}}{4} = 0 \\
&\therefore \left(x+\frac{c}{2}\right)^{2} + \left(y - \frac{d}{2}\right)^{2} = \frac{c^{2}+d^{2}}{4} - b \geq 0 \Longleftrightarrow c^{2} + d^{2} \geq 4b \Longleftrightarrow |a|^{2} \geq 4b
\end{align*}
