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?
There is an isomorphism between 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 isomorphism
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 the isomorphism.
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.