Sujet : Sign and complex.
De : r.hachel (at) *nospam* tiscali.fr (Richard Hachel)
Groupes : sci.mathDate : 03. Mar 2025, 22:37:07
Autres entêtes
Organisation : Nemoweb
Message-ID : <ULTAqmfHo9Brj6bDPiod8KaZGZg@jntp>
User-Agent : Nemo/1.0
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.