Find the conditions under which the given equation, in one complex unknown, has exactly one solution, and compute that solution The equation in question:
$az + b\bar{z}+c= 0$
The best i've managed so far, is to assume a and b are real numbers, but c is also a complex number. Then, solving for the real and imaginary parts of c, and re-arranging to find z, I wind up with:
$z = \frac{-Re(c) - Im(c)i}{a+b}$
The biggest flaw with this approach, seems to be the assumption that a and b are real. This also still seems to yield an infinite number of possible values for c , a, and b, which means an infinite number of solutions z?
 A: One can decompose $$z := x + i y$$ for $x, y \in \mathbb{R}$, and likewise decompose $a, b, c$ into real and imaginary parts. Then, collecting like terms gives a linear $2 \times 2$ system $$A \begin{pmatrix}x \\ y\end{pmatrix} = {\bf d}$$ in $x$ and $y$. It has a unique solution iff $$\det A \neq 0.$$

 In this case, $\det A = \pm(|a|^2 - |b|^2)$, that is, there is a unique solution iff $a$ and $b$ have different magnitudes.

A: Taking the conjugate of both sides in your equation,
$$\def\c#1{{\bar#1}}\c bz+\c a\c z+\c c=0\ .$$
Now consider the system of linear equations
$$\eqalign{
  az+bw&=-c\cr \c bz+\c aw&=-\c c\ .\cr}$$
If $a\c a-b\c b\ne0$ then this system has a unique solution, which can be found by standard linear algebra methods.  For your problem you also require $w=\c z$, but if you work out the solution you can confirm that this is in fact true.
Suppose on the other hand that $a\c a-b\c b=0$.  Then writing your equation as two real equations in $x,y$ gives a system in which the determinant is zero.  So once again, there is either no solution or infinitely many solutions.
