# Complex multiplication as rotation

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

• The fundamental connection between these is the formula $$e^{ix}=\cos x + i \sin x.$$ I suggest you to read "Visual Complex Analysis" by Tristan Needham for more information. – Kartik Nov 28 '15 at 13:53
• If you are not familiar with $e^{ix}$, you may consider the following (equivalent) explanation. – Ramiro Nov 28 '15 at 15:29
• Any complex numbers $a$ and $b$ can be written as $$a=|a|(\cos \alpha + i\sin \alpha)$$ and $$b=|b|(\cos \beta + i\sin \beta)$$ Using the rules of complex multiplication we get $$ab=|a||b|((\cos \alpha \cos \beta - \sin \alpha \sin \beta) + i(\cos \alpha \sin \beta + \cos \beta \sin \alpha))$$ But, we know that $$\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos (\alpha +\beta)$$ $$\cos \alpha \sin \beta + \cos \beta \sin \alpha = \sin(\alpha + \beta)$$ So $$ab=|a||b|((\cos (\alpha + \beta) + i \sin( \alpha + \beta))$$ And you see that adding angles means rotation. – Ramiro Nov 28 '15 at 15:30

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 $\theta$
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 $$zw = rse^{i\theta}e^{i\phi}=rse^{i(\theta+\phi)},$$ so as you can see the resulting complex number has angle $\theta+\phi$.