Show complex solutions exist

Solution 1:

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 $C$.

Solution 2:

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.