Complex multiplication as rotation

Question

Is there a reason that complex numbers multiplied so readily represent rotations in a plane? Any intuition behind this would help.

Answer 1

suppose you extend the field $\mathbb{R}$ by adjoining a root $\alpha$ of the equation: $$ x^2 + 1 = 0 $$ the extension field $\mathbb{R}(\alpha)$ properly contains $\mathbb{R}$ since an ordered field lacks square roots of elements less than zero.

$\mathbb{R}(\alpha)$ is a 2-dimensional vector space over $\mathbb{R}$, and we may take as a basis the pair $\{1,\alpha\}$. with respect to this basis this basis, multiplication by an element $c+\alpha d$ can be viewed as a linear transformation, with the matrix representation: $$ a+\alpha b \to \begin{pmatrix} a &-b \\ b &a \end{pmatrix} $$ the determinant $D=a^2+b^2$ is positive unless $a=b=0$. so we can find $\theta \in [0,2\pi)$ satisfying: $$ \cos \theta = aD^{-\frac12} \\ \sin \theta = bD^{-\frac12} $$ and the multiplication factorizes into a real multiplication coupled with an anticlockwise rotation through $\theta$

Answer 2

You can represent the complex number $z=x+iy$ by $z=re^{i\theta}$, where $r=\sqrt{x^2+y^2}$ and $\theta$ is the counter-clockwise angle from the positive $x$ axis. So if we multiply two complex numbers together, e.g. $z=re^{i\theta}$ and $w=se^{i\phi}$ we get $$zw = re^{i\theta}se^{i\phi}=(rs)e^{i(\theta+\phi)},$$ so as you can see the resulting complex number has angle $\theta+\phi$ and length $rs$.