I'm working throug Richard Earl's "Towards Higher Mathematics" and am stuck on this problem.
For what real numbers $A, B, C$ does the equation $$ z \bar z + Az + B \bar z = Ci $$ have a unique solution $z$?
What exactly does it mean for an equation to have a "unique" solution in this context? Would I be correct in assuming I merely substitute (a+bi) and (a-bi) and re-write as a quadratic in a or b and then use the discriminant? In that case would b^2 - 4ac = 0 or b^2 - 4ac > 0?
Thanks
With $z=x+iy$,
$$x^2+y^2+(A+B)x+i(A-B)y-Ci=0.$$
From the imaginary part, $y$ is unique when $$A\ne B.$$
Then
$$x^2+(A+B)x+y^2=0$$ requires $(A+B)^2=4y^2$ or
$$A+B=\pm2y=\pm\frac{2C}{A-B}$$
In other words,
$$A\ne B\land A^2-B^2=\pm2C.$$