Solvability conditions for $a z+ b\overline{z} = c$

Question

I would like to find solvability conditions for the linear algebraic equation

$$a z+ b\overline{z} = c$$

where $a, b, c \in \mathbb C$ are given. Can anyone help me?

Answer 1

If $z \in \Bbb C$ is a solution of $$ \tag{1} a z + b \bar z = c $$ then $$ \bar b z + \bar a \bar z = \bar c \, , $$ and "elimination" of $\bar z$ gives $$ \tag{2} (a \bar a - b \bar b) z = \bar a c - b \bar c \, . $$

Case 1: $|a| \ne |b|$. Then it follows from $(2)$ that $$ z = \frac{\bar a c - b \bar c}{|a|^2 - |b|^2} $$ and it is easily verified that this is really a solution of $(1)$. So we have exactly one solution in this case.

Case 2: $|a| = |b| > 0$. Define $\lambda := \sqrt{a/b}$, it does not matter which branch of the square root is chosen. Then $|\lambda| = 1$ and $(1)$ becomes $$ \lambda^2 b z + b \bar z = c \\ \Longleftrightarrow \lambda z + \bar\lambda \bar z = \frac{c}{\lambda b} \\ \Longleftrightarrow 2 \, \text{Re}(\lambda z) = \frac{c}{\lambda b} \, . $$ It follows that a solution exists if and only if $$ \frac{c}{\lambda b} = \frac{c}{\sqrt{ab}} \in \Bbb R \, , $$ and in that case all solutions are given by $$ z = \frac{1}{\lambda} \left( \frac{c}{2 \lambda b} + i t \right) = \frac{c}{2a} + i t \, \sqrt{\frac ba} \,, \quad t \in \Bbb R \, , $$ i.e. the solution set is a line.

Case 3: $a = b = 0$. That's an easy one ...

Answer 2

Let $z=x+iy$ then $\bar{z}=x-iy$

$$az+b\bar{z}=c$$ $$\Rightarrow (a_1+ia_2)(x+iy)+(b_1+ib_2)(x-iy)=c_1 + ic_2$$

Equating real and imaginary parts we have $$a_1x-a_2y+b_1x+b_2y=c_1 \Rightarrow (a_1+b_1)x+(-a_2+b_2)y=c_1$$ and $$a_2x+a_1y +b_2x-b_1y=c_2 \Rightarrow (a_2+b_2)x+(a_1-b_1)=c_2$$

For the above system to be solvable the determinant of coefficient matrix has to be non zero, i.e.,

$$(a_1+b_1)(a_1-b_1)-((-a_2+b_2)(a_2+b_2)) \neq 0$$ $$\Rightarrow {a_1}^2-{b_1}^2-{b_2}^2+{a_2}^2 \neq 0$$

This is same as $$|a| \neq |b|$$

Answer 3

if we show real parts with $r$_index and imaginary parts with $i$_index or in other words $a=a_r+ja_i , b=a_r+jb_i , c=c_r+jc_i,Z=x+jy$ where $j=\sqrt{-1}$ then we have: $$(a_r+ja_i)(x+jy)+(b_r+jb_i)(x-jy)=c_r+jc_i$$ equating real and imaginary parts gives us: $$ \begin{matrix} x(a_r+b_r)+ y(-a_i+b_i)& =c_r\\ x(a_i+b_i)+ y(a_r-b_r)& =c_i \end{matrix} $$ or in matrix form: $$\begin{pmatrix}a_r+b_r & -a_i+b_i\\ a_i+b_i & a_r-b_r \end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}=\begin{pmatrix}c_r\\ c_i \end{pmatrix}$$ this system is solvable if: $$\det\begin{pmatrix}a_r+b_r & -a_i+b_i\\ a_i+b_i & a_r-b_r \end{pmatrix}\neq0 \Rightarrow (a_r+b_r)(a_r-b_r)-(-a_i+b_i)(a_i+b_i) \neq 0$$
or in other words: $$a_r^2-b_r^2-(-a_i^2+b_i^2) \neq 0 \Rightarrow (a_r^2+a_i^2)-(b_r^2+b_i^2) \neq0 \Rightarrow |a|^2 \neq |b|^2 \Rightarrow |a| \neq |b| $$