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?
Best Answer
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.