How $a+bi$ becomes $\left(\matrix{a & -b\\b & a}\right)$? How do you change something into a matrix? E.g) 

How does $2+3i$ become $\left(\matrix{2 & -3\\3 & 2}\right)$

Is  there a specific rule for changing an input into a matrix that has to be learned? Or how is it done?
Thank you.
 A: It corresponds to one way to construct the field of complex numbers:
$$\mathbf C=\biggl\{\begin{pmatrix}a&-b\\b&a\end{pmatrix}\bigg\vert a,b\in\mathbf R\biggr\}=\biggl\{a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}0&-1\\1&0\end{pmatrix}\bigg\vert a,b\in\mathbf R\biggr\}$$
This is a commutative subring of $\mathcal M_2(\mathbf R)$.
In this construction, we identify the number $1$ with $I$, the unit $2\times 2$ matrix, a real number $a$ with the diagonal matrix $\begin{pmatrix}a&0\\0&a\end{pmatrix}$, and the imaginary number $\mathrm i$ with the matrix $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, which is justified by the fact that $J^2=-I$.
A: $$\begin{align}\color{blue}{(a+bi)\cdot}(x+yi)&=(ax-by)+(ay+bx)i\\
\color{blue}{\begin{bmatrix}a&-b\\b&a\end{bmatrix}\cdot}\begin{bmatrix}x\\y\end{bmatrix}&=\begin{bmatrix}ax-by\\ay+bx\end{bmatrix}
\end{align}$$
Do you see the connection? The correspondence between black items is maybe most clear; those are best thought of as elements. The blue items are best thought of as actions (multiplication by ...).
A: Fix a complex number $z=a+bi$ and consider the map $w \mapsto zw$. Considering $\mathbb C$ as $\mathbb R^2$, this is an $\mathbb R$-linear map whose matrix with respect to the canonical basis is the one given. Note that the canonical basis corresponds to $1, i \in \mathbb C$.
A: The idea is to create an isomorphism between the field of complex numbers, and a certain set of matrices that happen to form a field.
Specifically, complex numbers are a vector space of dimension $2$ over $\mathbb{R}$, with basis $\{1,i\}$. We would like a set of matrices that reproduce the addition (easy, just normal componentwise addition) and multiplication of complex numbers.
Addition, as mentioned above, is easy, and works for any matrices that can store two numbers. We want to multiply our matrices, so they had better be square. Now, either we can fiddle about to try and create some matrices that multiply correctly using Cartesian form, or we can take a shortcut by thinking about what multiplying complex numbers does geometrically.
In particular, the group action of multiplication on the left by a complex number, say $re^{i\theta}$, is isomorphic to the group of matrices acting as  rotations and scalings on points in the two-dimensional real vector space $\mathbb{R}^2$, using matrix composition. Let's call the first of these $(\mathbb{C}, \times)$, the second $(M,\cdot)$ Scalings are simply
$$ x \mapsto (rI)x= \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix}x, $$
while rotations are given by
$$ x \mapsto R(\theta)x = \begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix}x. $$
By taking real and imaginary parts and using Euler's formula $e^{i\theta} = \cos{\theta}+i\sin{\theta}$, it is a matter of a little algebra to show the composition laws for $(\mathbb{C}, \times)$ and $(M,\cdot)$ work in exactly the same way, that is, that $(re^{i\theta})\times(se^{i\phi}) = rse^{i(\theta+\phi)} $, and $(rR(\theta)) \cdot (sR(\phi)) = rsR(\theta+\phi) $.
Therefore if we define a map $f:\mathbb{C} \to M$ by
$$ re^{i\theta} \mapsto \begin{pmatrix} r\cos{\theta} & -r\sin{\theta} \\ r\sin{\theta} & r\cos{\theta} \end{pmatrix}, $$
we have $f(z+w) = f(z)+f(w)$ and $f(z \times w) = f(z) \cdot f(w)$. These are precisely the conditions for a field isomorphism. Therefore the two collections of objects act in the same way, and further, we can interpret the multiplication as rotation and scaling, which is why we chose a particular representitive of $i$ as $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, rather than choosing any old matrix square root of $-I$.
