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?

  • $\begingroup$ For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation. $\endgroup$ – dineshdileep Sep 19 '13 at 11:07
  • $\begingroup$ Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors. $\endgroup$ – Incnis Mrsi Nov 2 '14 at 7:09

There is a homeomorphism between the the complex numbers

$$ \color{blue}{a} + \color{red}{b}i $$

and the rotation matrices

$$ \left( \begin{array}{rc} \color{blue}{a} & \color{red}{b} \\ -\color{red}{b} & \color{blue}{a} \\ \end{array} \right) $$

Let the Cartesian forms be $$ z_{1} = \color{blue}{a} + \color{red}{b}i, \quad z_{2} = \color{blue}{c} + \color{red}{d}i $$ and the matrix forms $$ z_{1} = \left( \begin{array}{rc} \color{blue}{a} & \color{red}{b} \\ -\color{red}{b} & \color{blue}{c} \\ \end{array} \right), \quad z_{2} = \left( \begin{array}{rc} \color{blue}{c}& \color{red}{d} \\ - \color{red}{d} & \color{blue}{c}\\ \end{array} \right) $$


$$ \begin{align} % z_{1} + z_{2} &= (\color{blue}{a}+\color{blue}{c}) + (\color{red}{b} + \color{red}{d} )i \\[3pt] % &= \left( \begin{array}{rc} \color{blue}{a}+\color{blue}{c}& \color{red}{b}+ \color{red}{d} \\ -\color{red}{b}- \color{red}{d} & \color{blue}{a}+\color{blue}{c}\\ \end{array} \right) \end{align} $$


$$ \begin{align} z_{1} z_{2} &= (\color{blue}{ac}- \color{red}{bd}) + (\color{red}{b}\color{blue}{c}+\color{blue}{a}\color{red}{d})i\\[3pt] % &= % \left( \begin{array}{rc} \color{blue}{a}\color{blue}{c}- \color{red}{b} \color{red}{d} & \color{red}{b}\color{blue}{c}+\color{blue}{a} \color{red}{d} \\ - \color{red}{b} \color{blue}{c}-\color{blue}{a} \color{red}{d} & \color{blue}{a}\color{blue}{c}- \color{red}{b} \color{red}{d} \\ \end{array} \right) \end{align} $$


$$ \begin{align} % \frac{1}{z} &= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1} \left( \color{blue}{a} - \color{red}{b} i \right) \\[3pt] % &= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1} \left( \begin{array}{cr} \color{blue}{a} & - \color{red}{b} \\ \color{red}{b} & \color{blue}{a} \\ \end{array} \right) \end{align} $$

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    $\begingroup$ This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology. $\endgroup$ – Starfall Apr 24 '17 at 4:04
  • $\begingroup$ @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion. $\endgroup$ – dantopa Apr 24 '17 at 4:21

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