TITLE: Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$
QUESTION [43 upvotes]: 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.

REPLY [17 votes]: What I think of a complex number is a scaling and 2D rotation operation, where the absolute value $r$ is scaling factor, and the phase $\theta$ is the rotation angle.
The same operation can be described by scalar multiplication of a rotation matrix as $$r\begin{pmatrix}\cos \theta & -\sin \theta \\  \sin \theta & \cos \theta \end{pmatrix}$$
Since $r e^{i\theta}=r\cos \theta + ir \sin \theta = a +ib$, we have $$a +ib = \begin{pmatrix}a & -b \\  b & a \end{pmatrix}$$