# 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?

• 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. – dineshdileep Sep 19 '13 at 11:07
• 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. – 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}