The question asks me to show that "in general," $$z \overline{z} + az + \overline{b}z + c = 0$$ has two solutions, and asks when it has more than two.
Its obvious that if $a + b = 0$ and $c$ is negative, we get an equation of a circle.
Now, the two solutions i assume come from the fact that the equation looks very similar to a quadratic polynomial in $z$. I am just having trouble connecting $\|z\|^{2} + az + \overline{b}z + c = 0$ to $z^{2} + az + \overline{b}z +c = 0$
Thanks.
The issue of $a$ and $\overline{b}$ is not important; you really just have $|z|^2 + a z + c = 0$. Letting $a=a_1+ia_2,c=c_1+ic_2$, we have
$$x^2 + y^2 + a_1 x - a_2 y + c_1 + i (a_1 y + a_2 x + c_2) = 0$$
or
$$x^2 + y^2 + a_1 x - a_2 y + c_1 = 0, \\ a_1 y + a_2 x + c_2 = 0.$$
Complete the square in the first equation:
$$(x+a_1/2)^2 + (y-a_2/2)^2 + (c_1 - a_1^2/4 - a_2^2/4) = 0$$
So now our first equation has solution either a circle, one point, or no solutions. The second equation has solution of a line or the whole plane. The intersections are either a circle, two points, one point, or none. The "generic cases" (where we assume inequalities but no equalities) are two points or no points.