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?
 A: We have
$$
0=\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B =
z\overline{z}+\frac{1}{2}(Az+\overline{Az})+\frac{1}{4}\lvert A\rvert^2-\frac{1}{4}\lvert A\rvert^2+B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B
$$
If $4B>\lvert A\rvert^2>0$, then $\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B>0$, and hence no solutions.
If $4B>\lvert A\rvert^2\le 0$, set $C=\frac{1}{2}\sqrt{\lvert A\rvert^2-4B}$, and our equation is equivalent to
$$
\left\lvert z+\frac{1}{2}A\right\rvert^2=C^2,
$$
and hence equivalent to
$$
\left\lvert z+\frac{1}{2}A\right\rvert=C,
$$
the set of solutions of which is the circle centered at $-A/2$ with radius $C$.
A: 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.
A: 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
