Re: Sign and complex.

Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Complex numbers and products of different complex signs.
 What is a complex number?
 It is initially an imaginary number which is a duality.
 The two real roots of a quadratic curve, for example, are a duality.
 If we find as a root x'=2 and x"=4 we can include these two roots in a single expression: Z=3(+/-)i.
 Z is this dual number which will split into x'=3+i and x"=3-i.
 As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.
 Be careful with the signs (i=-1). If we add i, we subtract 1.
 If we subtract 5i, we add 5.
 But let's go further.
A small problem arises in the products of complexes.
 Certainly, if we take complexes of inverse spacings, that is to say (+ib) for one and (-ib) for the other, everything will go very well.
 Let's set z1=3-i and z2=4+2i.
 We have z1*z2=12+6i-4i-2i²=14+2i
 Let's set the inverse by permuting the signs of b:
 z1=3+i and z2=4-2i.
 We have z1*z2=12-6i+4i-2i²=14-2i
 We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.
 Question: Why does this formula become incorrect for complexes of the same sign in b?
 Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)
 The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has strictly NO content.
BTW, the result is correct in C, by the way C is defined. And it makes a lot of sense. And it can be *shown*.
Another formula could make sense too, in R(j) or R(epsilon) for instance.
But switching from a formula to another one depending of the sign of the imaginary part will lead to an inconsistent structure (i.e. il will not be an associative algebra on R, so it will be useless). BTW there are only three of them, see: https://www.youtube.com/watch?v=r5mccK8mNw8