How to deal with $\bar{x}$ when solving complex-variable linear equation(s) of x?

Question

The theory of linear algebra can be directly applied to linear equation(s) of complex variables with the form \begin{equation} \sum_i a_i x_i=c\ldots\ldots(1) \end{equation} with $a_i,c\in \mathbb{C}$.But I encounter difficulty when dealing with $$ \sum_i a_i x_i+\sum_j b_j \bar{x}_j=c\ldots\ldots(2) $$ $x_i$ and $\bar{x_i}$ are not independent so the latter cannot be thought as a completely new variable. However I cannot see any simple relationship between $x_i$ and $\bar{x}_i$ so I cannot replace $\bar{x}_i$. The only way I have in mind is to write every complex variable $x_i$ as linear combination of two real variables $x_i=x_{i,real}+ix_{i,img}$ and convert the original equation(s) into real-variable equations with twice the number.

Here is my question:

How can we convert complex-variable equation(s) with the form of Eq(2) into equations with the form of Eq(1) that can be readily solved by the theory of linear algebra?

Answer 1

There are two ways. One, which is what you thought of, is to replace $x_k$ with $a_k+ib_k$ where $a_k,b_k$ are real. (By the way, it's best to avoid $i$ as an index in this context.) Then separate the real and imaginary part in every equation. Now you have twice as many variables and twice as many equations as you started with. But everything is over the real scalars, and all $a_k,b_k$ are independent variables.

Another, which takes less re-writing, is to conjugate every equation and add the conjugated forms to the system. Then solve it, treating $x_i$ and $\bar x_i$ as independent variables. Magically, the solution will be such that the value of $\bar x_i$ is the conjugate of the value of $x_i$.

Random example: let's solve
$$\begin{split} z+(1-i)w +2i\bar z - \bar w &= 4+2i \\ iz + (2+3i)w - 2\bar z + 3i \bar w &= 3-i \end{split} $$ The conjugated equations are $$\begin{split} \bar z+(1+i)\bar w - 2i z - w &= 4 -2i \\ -i \bar z + (2-3i)\bar w - 2 z - 3i w &= 3+i \end{split} $$ In matrix form, the entire $4\times 4$ system is $$\begin{pmatrix} 1 & 1-i & 2i & -1 \\ i & 2+3i & -2 & 3i \\ -2i & -1 & 1 & 1+i \\ -2 & -3i & -i & 2-3i \end{pmatrix} \vec x = \begin{pmatrix} 4+2i \\3-i \\ 4-2i \\3+i \end{pmatrix} $$ This is easier to type than it looks, because a lot of it is copy-paste-conjugate. Notice in particular the block structure of the matrix on the left. Solution (found numerically, with A\b in Scilab): $$ \begin{pmatrix} - 4.0909091 + 2.4545455\,i \\ - 1.3636364 + 3.1818182\,i \\ - 4.0909091 - 2.4545455\,i \\ - 1.3636364 - 3.1818182\,i \end{pmatrix}$$ The top two rows are $z,w$; the bottom two are their conjugates.