Solving a complex equation $Ax=b$ with $A,b\in\mathbb C$ using linear algebra Consider the following very simple linear system with one unknown:
\begin{equation}\label{a}\tag{1}
Ax=b \\
\left ( 3+4i \right )x=(6+8i).
\end{equation}
This paper ("On the numerical solving of complex linear systems") says that I can solve the linear system by transforming A to matrix form and then solving it as follows:
\begin{equation}\label{b}\tag{2}
\begin{pmatrix}
3 & -4\\ 
4 & 3
\end{pmatrix}
\binom{x_r}{x_c}
=
\binom{b_r}{b_c},
\end{equation}
where $b_r = 6,b_c=8$.
Question: The translation of the A is fairly easy to understand. What I don't get is why b is not converted to matrix form yet solving the above system yields the correct answer. In other words, the following equations
\begin{equation}\label{c}\tag{3}
\left\{\begin{matrix}
3x_r-4x_c = 6\\ 
4x_r+3x_c = 8
\end{matrix}\right.
\end{equation}
makes no sense to me. 
Note that, I know how to solve the linear system. I'm looking for a detailed explanation of what's happening between (1) and (2)
Thank you
 A: Let's try to justify the equations given by (3).  If we write out the product $(3+4i)x$, we have
$$
(3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \\
= [3x_r - 4x_c] + [4x_r + 3x_c]i
$$
Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.  So, in order to have $(3+4i)x = b$, we must have
$$
3x_r - 4x_c = b_r\\
4x_r + 3x_c = b_c
$$
which is precisely the system of equations that you've come up with.
A: Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.
$$\begin{bmatrix}3&-4\\4&3\end{bmatrix}\begin{bmatrix}x_r&-x_c\\x_c&x_r\end{bmatrix}=\begin{bmatrix}b_r&-b_c\\b_c&b_r\end{bmatrix}$$
It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.

As an example where the simplified version will fail we can take a look at permutations. A permutation among three different elements can be represented with binary matrices:
$$P_1 = \left[\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right], P_2 = \left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]$$
Now let us say that if we apply $P_1$ to some permutation and it becomes $P_2$, then what was the original permutation?
This we can express as: $$P_1X = P_2$$
Here we need $X$ to be a 3x3 matrix, because there is no other way to represent it.
The answer is: $$X=P_1 ^{-1}P_2$$
A: $x=x_r+ix_i$. When we multiply with $3+4i$ we get $3x_r+3ix_i+4ix_r-4x_i=6+8i$. We now equate the real and imaginary parts separately:$$3x_r-4x_i=6\\3x_i+4x_r=8$$
A: You want to solve $az=b$ for $a,b\in\mathbb C$, using the machinery of linear algebra.
Denote real and imaginary parts of the quantities as $z=x+iy$ and $a=a_R + i a_I, b= b_R+i b_I$. The equation is then equivalent to the following system of real equations:
$$\begin{cases}
  a_R x - a_I y = b_R, \\
  a_R y + a_I x = b_I,
\end{cases}$$
which can be written in matrix form as
$$\underbrace{\begin{pmatrix}
  a_R & -a_I \\
  a_I & a_R
\end{pmatrix}}_{\equiv A}
\begin{pmatrix} x \\ y \end{pmatrix} =
\begin{pmatrix} b_R \\ b_I \end{pmatrix}.
$$
Note that the determinant of the matrix $A$ equals $|a|^2$, which is consistent with $az=b$ having nontrivial solutions iff $a\neq0$.
Assuming $|a|^2\neq0$, we can invert $A$ obtaining
$$\begin{pmatrix} x \\ y \end{pmatrix} =
\frac{1}{|a|^2}\begin{pmatrix} a_R & a_I \\ - a_I & a_R \end{pmatrix}
\begin{pmatrix} b_R \\ b_I \end{pmatrix},$$
which translated back into a linear system is
$$\begin{cases}
  |a|^2 x = a_R b_R + a_I b_I, \\
  |a|^2 y = -a_I b_R + a_R b_I,
\end{cases}$$
or summing the first equation to the second times $i$, gives
$$z\equiv x+iy = \frac{1}{|a|^2}[(a_R b_R + a_I b_I) + i (-a_I b_R+a_R b_I)].$$
To see that this is consistent with the solution we get via complex analysis, $z=b/a$, write the latter as $z=ba^*/|a|^2$, and verify that the two expressions are equal.
