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?

Let $z=x+iy, A=a+ib$  where $x,y,a,b$ are real
So, we have $x^2+y^2+ax-by+B=0$
$\implies\left(x+\dfrac a2\right)^2+\left(y-\dfrac b2\right)^2=B-\dfrac{a^2+b^2}4$
But $\left(x+\dfrac a2\right)^2+\left(y-\dfrac b2\right)^2\ge0$
Hope you can take it from here
