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.

I had something written up on this lying around. The $-$ sign is off, but it's more or less the same, I hope it helps.
Let $M$ denote the set of such matrices. Define a function $\phi\colon M\to\mathbb{C}$ by
$$
\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha\end{pmatrix}\mapsto \alpha+i\beta.
$$
Note that this function has inverse $\phi^{-1}$ defined by $\alpha+i\beta\mapsto\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha\end{pmatrix}$. This function is well defined, since $\alpha+i\beta=\gamma+i\delta$ if and only if $\alpha=\gamma$ and $\beta=\delta$, and thus it is never the case that a complex number can be written in two distinct ways with different real part and different imaginary part. So $\phi$ is invertible.
Now let 
$$
A=\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha\end{pmatrix},\qquad 
B=\begin{pmatrix} \gamma & \delta \\ -\delta & \gamma\end{pmatrix}.
$$
Then
$$
\phi(A+B)=\phi\begin{pmatrix} \alpha+\gamma & \beta+\delta \\ -\beta-\delta & \alpha+\delta\end{pmatrix}=(\alpha+\gamma)+i(\beta+\delta)=(\alpha+i\beta)+(\gamma+i\delta)=\phi(A)+\phi(B).
$$
Also,
$$
\phi(AB)=\phi\begin{pmatrix} \alpha\gamma-\beta\delta & \alpha\delta+\beta\gamma \\ -\beta\gamma-\alpha-\delta & -\beta\delta+\alpha\gamma\end{pmatrix}=(\alpha\gamma-\beta\delta)+i(\alpha\delta+\beta\gamma)=(\alpha+i\beta)(\gamma+i\delta)=\phi(A)\phi(B).
$$
So $\phi$ respects addition and multiplication. Lastly, $\phi(I_2)=1$, so $\phi$ also respects the multiplicative identity. Hence $\phi$ is a field isomorphism, so $M$ and $\mathbb{C}$ are isomorphic as fields.
