Find the conditions under which the equation $a z + b \overline{z} + c = 0$ in one complex unknown has exactly one solution, and compute that solution.
In the following solution that I found to this problem, I'm unsure how they rewrote $a(x+iy)+b(x-iy)+c=0$ as equations $(1.6a)$ and $(1.6b)$. Why do we get two equations? And I'm just starting to read Ahlfors "Complex Analysis" and this is a question early in the text.
Thanks
Let $z = x + iy$. Then $az + b\overline{z} + c = a(x+iy) + b(x-iy) + c = 0$. \begin{align} (a+b)x + c &= 0 \tag{1.6a} \\ (a-b)y &= 0 \tag{1.6b} \end{align} Lets consider equation $(1.6b)$. We either have that $a = b$ or $y = 0$. If $a = b$ then WLOG equation (1.6a) can be written as $$ x = \frac{-c}{2a} $$ and $y \in \mathbb{R}$. For fixed $a,b,c$, we have infinitely many solutions when $a = b$ since $z = -c/(2a) + iy$ for $y \in \mathbb{R}$. […]
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This is just the basic fact about complex numbers. For complex numbers $u,v$:
$$ u = v \qquad \text{if and only if} \qquad \mathrm{Re}(u) = \mathrm{Re}(v) \quad \text{and} \quad \mathrm{Im}(u) = \mathrm{Im}(v) $$
Equations 1.6a and 1.6b are just the real and imaginary parts of the original equation to solve (assuming $a,b,c,x,y$ are all real numbers), and so together they are equivalent to the original equation.