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

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    $\begingroup$ 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. $\endgroup$ – Kartik Nov 28 '15 at 13:53
  • $\begingroup$ If you are not familiar with $e^{ix}$, you may consider the following (equivalent) explanation. $\endgroup$ – Ramiro Nov 28 '15 at 15:29
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    $\begingroup$ 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. $\endgroup$ – Ramiro Nov 28 '15 at 15:30
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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 $\theta$

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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 $\theta+\phi$.

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