# Do three value complex numbers exist? [duplicate]

My understanding of complex numbers is that it is the use of an extra axis of dimension. By plotting a signal on polar axis, the real number, x, is the sine of the signal, and the imaginary number, y, is the cosine of the signal. This understanding is derived from its use in the fourier transform.

(By this understanding, one could take the cosine of a value to get its imaginary component, so my understanding it limited as it doesn't explain the logic behind √(−1)=i.)

So, I have the question, is it possible to have three value complex numbers?

You can have 1, 2, 4, and 8-dimensional real normed division algebras.

What does that mean? It means that if you want to find a way to multiply lists of $$n$$ numbers (or $$n$$-dimensional vectors, alternatively), you need $$n=1,2,4,$$ or $$8$$ to get a meaningful answer. It is difficult to say why only these numbers work. As one elementary indication that perhaps 3-dimensions is spooky, note that multiplication of complex numbers multiplies length; that is,

$$|a+bi| \cdot |c+di| = |(a+bi) \cdot (c+di)|.$$

The length of the product is the product of the lengths. This leads to an identity

$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2,$$

where a product of two sums of two squares is also a sum of two squares.

However, a product of a sum of three squares is not always a sum of three squares, so no such identity exists for triples.

• Interesting, so the perspective of it being an additional axis is only applicable to the 1D fourier transform. Nov 1, 2020 at 18:42