Complex numbers and their matrix form. I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane.  This thing can be called a vector and be represented as $(a,b), [a\text{ }b]^T$ (column vector) or by $a\mathbf{i}+b\mathbf{j}$, where $(\mathbf{i},\mathbf{j})$ is the stardard basis in $\mathbb{R}^2$  Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=\sqrt{-1}$.
So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations 
$\begin{bmatrix} \cos(90) & -\sin(90) \\ \sin(90) & \cos(90)\\ \end{bmatrix}$=$\begin{bmatrix} 0 & -1 \\ 1 & 0\\ \end{bmatrix}$ and we find that
$$\begin{bmatrix} 0 & -1 \\ 1 & 0\\ \end{bmatrix}\begin{bmatrix} a \\ b\\ \end{bmatrix}=\begin{bmatrix} -b \\ a\\ \end{bmatrix}$$
Or, I could choose the complex multiplication way and say, 
$i(a+bi)=ai+bi^2=ai-b=-b+ai$
So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?
 A: There is a homeomorphism between the the complex numbers
$$
\color{blue}{a} + \color{red}{b}i
$$
and the rotation matrices
$$
\left(
\begin{array}{rc}
 \color{blue}{a} & \color{red}{b} \\
 -\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)
$$
where $\color{blue}{a}^{2} + \color{red}{b}^{2} = 1.$ We see the familiar rotation matrix
$$
R(\theta) = 
\left(
\begin{array}{rc}
 \cos \theta & \sin \theta \\
 -\sin \theta & \cos \theta \\
\end{array}
\right)
$$
which in the form
$$
x'=R(\theta)x
$$
rotates the $2-$vector $x$ about the origin by $\theta$, producing the $2-$vector $x'$.

Verify homeomorphism
Start with two complex numbers $z_{1}$ and $z_{1}$. The Cartesian forms are
$$
 z_{1} = \color{blue}{a} + \color{red}{b}i, \quad 
 z_{2} = \color{blue}{c} + \color{red}{d}i
$$
where the numbers $a$, $b$, $c$, and $d$, are all real. Blue numbers signify the real component of $z$, and red the imaginary component.
Equivalent matrix forms are defined as
$$
z_{1} = \left(
\begin{array}{rc}
 \color{blue}{a} & \color{red}{b} \\
 -\color{red}{b} & \color{blue}{c} \\
\end{array}
\right),
\quad 
z_{2} = \left(
\begin{array}{rc}
 \color{blue}{c}& \color{red}{d} \\
 - \color{red}{d} & \color{blue}{c}\\
\end{array}
\right)
$$
Verify basic properties of of the homeomorphism.
Addition
$$
%
z_{1} +
z_{2} =
(\color{blue}{a} + \color{red}{b}i) +
(\color{blue}{c} + \color{red}{d}i) 
=
(\color{blue}{a}+\color{blue}{c}) + (\color{red}{b} + \color{red}{d} )i 
$$
$$
%
z_{1} +
z_{2} =
\left(
\begin{array}{rc}
 \color{blue}{a} & \color{red}{b} \\
 -\color{red}{b} & \color{blue}{c} \\
\end{array}
\right) +
\left(
\begin{array}{rc}
 \color{blue}{c} & \color{red}{d} \\
 -\color{red}{d} & \color{blue}{c} \\
\end{array}
\right) 
= \left(
\begin{array}{rc}
 \color{blue}{a}+\color{blue}{c}& \color{red}{b}+ \color{red}{d} \\
 -\color{red}{b}- \color{red}{d} & \color{blue}{a}+\color{blue}{c}\\
\end{array}
\right)
$$
Multiplication
$$
z_{1} z_{2} =
(\color{blue}{a} + \color{red}{b}i) 
(\color{blue}{c} + \color{red}{d}i) 
=
(\color{blue}{ac}- \color{red}{bd}) + 
(\color{red}{b}\color{blue}{c}+\color{blue}{a}\color{red}{d})i
%
$$
$$
%
z_{1} z_{2} =
\left(
\begin{array}{rc}
 \color{blue}{a} & \color{red}{b} \\
 -\color{red}{b} & \color{blue}{c} \\
\end{array}
\right) 
\left(
\begin{array}{rc}
 \color{blue}{c} & \color{red}{d} \\
 -\color{red}{d} & \color{blue}{c} \\
\end{array}
\right) 
=
%
\left(
\begin{array}{rc}
 \color{blue}{a}\color{blue}{c}- \color{red}{b}  \color{red}{d} & \color{red}{b}\color{blue}{c}+\color{blue}{a} \color{red}{d} \\
 - \color{red}{b} \color{blue}{c}-\color{blue}{a} \color{red}{d} & \color{blue}{a}\color{blue}{c}- \color{red}{b}  \color{red}{d} \\
\end{array}
\right)
$$
Inversion
$$
%
\frac{1}{z} 
= \frac{1}{\color{blue}{a} + \color{red}{b} i} 
= \left( \frac{\color{blue}{a} - \color{red}{b} i}{\color{blue}{a} - \color{red}{b} i} \right)
\frac{1}{\color{blue}{a} + \color{red}{b} i} 
= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1} \left( \color{blue}{a} - \color{red}{b} i \right) %
$$
$$
z^{-1} 
= \left(
\begin{array}{cr}
 \color{blue}{a} & - \color{red}{b}  \\
 \color{red}{b} & \color{blue}{a} \\
\end{array}
\right)^{-1} 
= \frac{\text{adj }z}{\det z}
= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1}
\left(
\begin{array}{cr}
 \color{blue}{a} & - \color{red}{b}  \\
 \color{red}{b} & \color{blue}{a} \\
\end{array}
\right)
$$
where adj $z$ is the adjugate matrix.
