Re: Equation complexe

Le 26/02/2025 à 13:31, Moebius a écrit :

If x ∈ IR, then x^2 ∈ IR and x^2 >= 0, and hence x^4 = (x^2)^2 >= 0.
 So x can't be in IR.
 Now let x = ir with r ∈ IR (and i, the imaginary unit, ∈ C).
 Then x^2 = i^2 * r^2 = -1 * r^2 ∈ IR. Hence x^4 = (x^2)^2 >= 0.
 Hence neither x ∈ IR nor x = ir with r ∈ IR will work.
 Actually, any x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2,  (-1-𝑖)⋅3/√2} will work.
 Hope this helps.

Thank you.
But we are not talking about the same thing.
Mathematicians talk about an imaginary unit i.
I also talk about an imaginary unit i.
Mathematicians propose to replace in a negative discriminant a multiplication by 1, by a multiplication by -i², since i²=-1 and 1=-i².
I propose strictly the same thing, and, obviously, I obtain the same results as them.
That is not the problem.
Simply, mathematicians do not explain WHY their imaginary unit i is worth -1 when we square it. They make dictates. Useful dictates, interesting dictates, but dictates nonetheless.
They don't seem to realize what this imaginary unit is that they use, and only see it under a distant i²=-1 that is very limited and that explains nothing, and that plunges them into error as soon as they use anything other than simple quadratic equations.
If we ask them to solve a degree 4 equation without real roots, for example, everything will collapse in horror, and they will come up with absolutely anything.
Let's set f(x)=x^4+(9/4) or f(x)=x^4+x²+2 and the complex roots that they give are so ridiculous that they make you laugh.
This shows that they have not understood what i was and how to correctly handle this unit in basic analytics. I am not talking about the Argand frame, that is something else. I am talking about the role of i in the Cartesian frames of reference. The Argand frame is "something else" where the modulus of a complex is defined as the square root of the product of its two conjugate complexes.
R.H.