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Let $(a,b)$ and $(c,d)$ be vectors in $\mathbb{R}^{2}$. Define a function $\langle\:,\:\rangle\colon\mathbb{R}^{2}\times\mathbb{R}^{2}\to\mathbb{R}$ putting \begin{equation} \langle (a,b),(c,d)\rangle=ad+bc. \end{equation} One easily checks that this is a bilinear form which is symmetric and nondegenerate.

It is not positive-definite in general but I think we still can talk about orthogonality without any loss. We say vectors $(a,b)$ and $(c,d)$ are orthogonal if $\langle (a,b),(c,d)\rangle=0$.

Now, if we identity $\mathbb{R}^{2}$ with the complex numbers $\mathbb{C}$, then one readily identifies the expression $ad+bc$ as the imaginary part in the product $(a,b)\cdot (c,d)$ (or $(a+bi)(c+di)$ if you like).

In this sense, the above bilinear form has the cool slogan

Complex numbers are orthogonal if and only if their product is real".

This slogan sounds so good and this probably means that this function is already known and has some importance in some area. This question asks for references or any tips at all on this.

Not that is relevant, but this investigation arose from some number-theoretic mess-arounds in the field $\mathbb{Q}(i)$, so if this shows up in this area it would be a could thing to know. Thanks

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    This is one of the standard forms for a Lorentzian inner product on $\Bbb R^2$, chosen so that the null cone is just the union of the coordinate axes. – 2017-02-22
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    The interesting thing is that if we consider a group ring $kG$ for $k=\mathbb{R}$ and $G=\mathbb{Z}_2$ and we identify $kG=\mathbb{C}$ with the basis $G=\{1, i\}$ then the imaginary part of $z*z'$ (with group ring multiplication) is exactly $$. Not sure if this is of relevance though. – 2017-02-22

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