Re: Equation complexe

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Sujet : Re: Equation complexe
De : dohduhdah (at) *nospam* yahoo.com (sobriquet)
Groupes : sci.math
Date : 28. Feb 2025, 02:40:09
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Op 28/02/2025 om 01:50 schreef Richard Hachel:
Le 28/02/2025 à 00:53, "Chris M. Thomasson" a écrit :
Claro que si.
Pero, i*i=i²=-1 ; (i²)²=-1
It seems that the imaginary unit i is a special unit such that i^x=-1 whatever x.
>
Mathematicians are right when they say that i=-1, that i^(-1/2)=-1, that i²=-1.
But if we understand Dr. Hachel's idea, we see that these three true statements are not enough.
Hachel imposes that i is an imaginary unit such that i^x=-1 whatever x.
This confuses the mathematician, who is used to working with real numbers, and who sets a²*a²=a^4 with systematically a>0.
But here we are not working with real numbers, but with the imaginary i.
It is not the same thing: we must systematically set i^x=-1 for all x.
>
It kind of seems like you deny that the y axis even exists?
 Where do you see that I deny the existence of the y-axis?
I do not deny the existence of the y-axis, on the contrary, I affirm that in a Cartesian coordinate system, there are two axes, and that we can draw a coordinate system Ox, Oy.
In this coordinate system, we can draw straight lines, curves, etc...
We can draw, for example f(x)=2x+1, or g(x)=sqrt(x)+2, or h(x)=x²+2x+1.
Now, we can look for roots, that is to say the place where these functions cross the y=0 axis. If there are no roots, we can sometimes look for complex roots.
Simply, in Hachel, all real or complex roots must be on the x'Ox axis, that is to say on the y=0 line.
Trying to place roots "elsewhere" is particularly absurd, since by definition, the place where a function crosses x'Ox is on x'Ox.
We will then say, but if the function does not cross x'Ox, what happens, like for example f(x)=x²+4x+5 which has no real roots. We must then rotate the curve in such a way that roots will appear, which will be the real roots of the mirror curve, and at the same time the complex roots of the original curve.
Thus, the complex roots of a curve are the real roots of the mirror curve, and vice versa.
We will find here that f(x) has two complex roots which are x'=3i, and x"=i which are the same points as x'=-3 and x'=-1 but noted differently, depending on whether we indicate that they are the real roots of g(x) or the complex roots of f(x).
The points A and B thus noted on the Cartesian coordinate system are A(-3,0) and B(-1,0), or, in mirror, complex coordinates, A(3i,0) and B(i,0).
 Thus we note that the complex axis IS the x'Ox axis, but inverted, and that it only concerns the abscissas. The y-axis remains in its place, and is used to position the ordinate.
 Then, we can resort, for other reasons, to an Argand-Gauss coordinate system, which we use in a completely different way (orthogonalization of x in a+ib), but these are two very different things, and two totally different reference points that should not be confused and even less used as confused on the same diagram.
 R.H.
Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2.
https://youtu.be/GZegwJVC_Pc?t=67
You claim that -1 * -1 = -1
since you claim that i^4 = -1 = i^2 * i^2 = -1 * -1

Date Sujet#  Auteur
25 Feb 25 * Equation complexe101Richard Hachel
25 Feb 25 +* Re: Equation complexe73Jim Burns
25 Feb 25 i`* Re: Equation complexe72Richard Hachel
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28 Feb 25 i   i       `* Re: Equation complexe19Chris M. Thomasson
28 Feb 25 i   i        +* Re: Equation complexe17Richard Hachel
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26 Feb 25 i   i `- Re: Equation complexe1Chris M. Thomasson
26 Feb 25 i   +* Re: Equation complexe3Chris M. Thomasson
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26 Feb 25 i   `* Re: Equation complexe5Barry Schwarz
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