Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$ 
Possible Duplicate:
Relation of this antisymmetric matrix $r = \!\left(\begin{smallmatrix}0 &1\\-1 & 0\end{smallmatrix}\right)$ to $i$

On Wikipedia, it says that:

Matrix representation of complex numbers
Complex numbers $z=a+ib$ can also be represented by $2\times2$ matrices that have the following form: $$\pmatrix{a&-b\\b&a}$$

I don't understand why they can be represented by these matrices or where these matrices come from.

As to the "where did it come from?", rather than verifying that it does work: this is a special case of a "rational representation" of a bigger collection of "numbers" as matrices with entries in a smaller collection. ("Fields" or "rings", properly, but it's not clear what our context here is.)
That is, the collection of complex numbers is a two-dimensional real vector space, and multiplication by $a+bi$ is a real-linear map of $\mathbb C$ to itself, so, with respect to any $\mathbb R$-basis of $\mathbb C$, there'll be a corresponding matrix. For example, with $\mathbb R$-basis $e_1=1,\,e_2=i$, 
$$
(a+bi)\cdot e_1 = a+bi = ae_1+be_2
\hskip40pt
(a+bi)\cdot e_2 = (a+bi)i = -b+ai = -be_1+ae_2
$$
So
$$
\pmatrix{e_1 \cr e_2}\cdot (a+bi) \;=\; \pmatrix{a & b \cr -b & a}\pmatrix{e_1\cr e_2}
$$
Oop, I guessed wrong, and got the $b$ and $-b$ interchanged. Maybe using $e_2=-i$ instead will work... :)
But this is the way one finds such representations.
