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 $\theta$