-2

Every complex number axiom system, that I have ever seen, contains one of these two statements: 1) Exists a complex number $i$ : $i^2 = -1$; or 2) For any complex numbers $z_1=a+bi, z_2=c+di$ their product $z_1 \cdot z_2=(ac-bd)+(ab+cd)i$.

All that looks very unreasonable, taken from nowhere. I can't understand, why so strangely constructed set generalizes the real numbers and even keeps linearity, just as vectors (i.e. $(a+bi)+(c+di)=(a+b)+(c+d)i$ and $c(a+bi)=(ca)+(cb)i$). Of course, it's not just a coincidence!

What stands behind all this stuff and where does it come from? Why are complex numbers usually defined like that and what is a more clear way?

P.S. The question is not about how do complex numbers work, but why do they work and why does their particular model describe the most natural extension of real numbers.

Andrew Dmitrenko
  • 7
  • 1
    (2) is taken from arranging what $(a+bi)(c+di)$ becomes if you assume simultaneously: (a) commutativity and associativity of sum an product, (b) distributivity of product over sum and (c) $i\cdot i=-1$ –  Sep 01 '20 at 18:47
  • 1
    Does This explanation help? – Mark Viola Sep 01 '20 at 18:49
  • I think it would make more sense to you if we construct the complex numbers by the relation $$\mathbb{C}=\mathbb{R}[x] / \langle x^2 + 1 \rangle$$see also this question. – Peter Foreman Sep 01 '20 at 18:49
  • 3
    "taken from nowhere. I can't understand,..." Remember, the problems we want to solve come first. The definitions are explicitly written in such a way as to have the desired properties and to be useful for those problems we wanted solved. – JMoravitz Sep 01 '20 at 18:52
  • @Gae.S. Unfortunately these models still assume, that $i^2=-1$, what disturbes me so much. – Andrew Dmitrenko Sep 01 '20 at 18:57
  • @MarkViola Well, it bases on assumption that $i^2=-1$, but I want do without it. – Andrew Dmitrenko Sep 01 '20 at 19:02
  • 1
    Why should it disturb you so? Someone may have once said "Oh, you know what would be great? If there was actually a number such that $z^2 = -1$. Wouldn't that be neat? Because if there were such a number then we could do all this fancy algebra and find the roots of these cubic polynomials and we wouldn't have to be so frustrated when we found negative numbers under square roots and cube roots... Hmm... well, in order to have such a number we need to have such and such properties defined..." – JMoravitz Sep 01 '20 at 19:03
  • Basing a definition around a desirable property which is clearly a useful property to have is not a bad thing! – JMoravitz Sep 01 '20 at 19:06
  • @JMoravitz I know, classical definition is very useful and helps to solve many different problems. But I want to look deep into it, learn the fundamantals. It's normal to think about real numbers, like just an instrument to count some things, but as for me, it's looks much better, when you think about real numbers, as a continued ordered linear space! Same with complex numbers: yes, it's useful instrument, but how and why does it work? - that's the main question. – Andrew Dmitrenko Sep 01 '20 at 19:13
  • It works like the splitting field of $x^2+1$ over the reals. Oh wait; that means we adjoins an element $x$ with $x^2=-1$ to the real numbers. – saulspatz Sep 01 '20 at 19:16
  • Does this answer your question? What are imaginary numbers? – rschwieb Sep 01 '20 at 20:22
  • This sounds like every other "how to complex numbers work" question we have, I think. – rschwieb Sep 01 '20 at 20:23
  • Well, now I realize, that it's quite the same about thinking about roots of the equation $x^2+1=0$ as about roots of the equation $x+1=0$ (negative numbers) or $2x-1=0$ (fractions) or $x^2-2=0$ (irrational numbers). I think, to understand it deeper I should learn something about field extensions in general. But for now it's enough. Thank you all for the conversation! – Andrew Dmitrenko Sep 02 '20 at 14:17
  • @PeterForeman So I consider your idea most natural. Thank you! – Andrew Dmitrenko Sep 02 '20 at 14:25

1 Answers1

3

I'm very curious about your use of the word "unreasonable". When your teacher told you that you were going to start using a number that you would write as $\sqrt 2$, and that

$$ (1 + 5 \sqrt 2) + (3 - 4 \sqrt 2) = (1+3) + (5-4)\sqrt 2 = 4 + 1\sqrt 2,$$

just how reasonable / unreasonable did you find that?

And to take it one step further, what do you think of the way that $ (1 + 5 \sqrt 2) \times (3 - 4 \sqrt 2)$ was calculated?

JonathanZ
  • 10,615