Show complex solutions exist Let A be a complex number and B a real number. Show that the equation 
$\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,$ has a solution iff $\,\lvert A^2\rvert \geq 4B$. If this is so, show that the solution set is a circle or a single point.
Well i am trying to do the first part first. So assuming the equation has a solution that would mean $z = x+iy$ satisfies the equation. 
I was going to let $A = s+it$ for a complex number, but it is not working out for me. Wrong step?

You are on the right track by letting $z = x+iy$ and $A=s+it$.  If substitute this in, we have
$$
x^2+y^2+xs - yt+r = 0
$$
where $r$ is my real number.
Try completing the square for $x$ and $y$ and see what you get.
