Fix $A \in ℂ$ and $B \in ℝ$ Let $z \in ℂ$.
Show that the equation $|z^2| + Re(Az) + B = 0$ has solutions iff $|A^2| ≥ 4B$
I have no trouble proving the forward implication, but its the "only if" part that I can't prove. I am writing the complex numbers in polar form and then working from there with the discriminant. As far as I can tell, the $|A^2| ≥ 4B$ is not a sufficient condition to imply that the equation will have solutions. Can someone help me to see what I am doing wrong?