Re: New equation

Le 26/02/2025 à 01:05, "Chris M. Thomasson" a écrit :

On 2/25/2025 3:58 PM, Richard Hachel wrote:

Le 26/02/2025 à 00:03, "Chris M. Thomasson" a écrit :

>

 You don't understand what I'm saying.
But that's okay.
When I use a Cartesian coordinate system, whether in two or three dimensions, I use two or three real axes.
Ox,Oy,Oz.
So far, so good, everyone understands.
Let's just go back, breathe, blow, to a Cartesian plane, which is very simple.
I place my "x" on the abscissa, and my "y" on the ordinate.
And finally, I draw my curves...
I draw the curve f(x)=x²+4x+5.
I've been told that this is colossally difficult, and that given the level of the participants in sci.maths, who are very stupid and barely know how to draw the line y=2x+1, I shouldn't be talking about curves, and even less about imaginary numbers.
But I am naturally optimistic, I tell myself that, perhaps, on sci.math there are intelligent people, more intelligent than the average French person.
So I will draw my curve, and, surprise! No roots.
So I cannot say that there is a root at A(2,0) and another at B(5,0), since there is none. There is none.
I repeat (given the stupidity of human beings in general, I have to repeat often), the equation f(x)=x²+4x+5 has no root, and I cannot draw anything at all on my x'Ox axis.
That is when I realize that, by mirror effect, if I place another mirror curve that touches the first at the top, my curve will cross my axis at two points.
Breathe, blow...
This imaginary curve, which is not f(x), I'm going to call it g(x), and I'm going to give it an equation. g(x)=-x²-4x-3. And there, I'm going to give it two roots. x'=-3 and x"=-1.
So f(x) has no real roots, but two imaginary roots on its mirror curve, and g(x) has no imaginary roots, but two real roots.
That said, I cannot grant the real roots of g(x) to f(x), but I can attribute imaginary mirror roots to it via g(x).
Simply I cannot say that the real roots of f(x) are x'=-3 and x"=-1, I have to say that they are its imaginary roots of the mirror curve, and to specify it well, it is necessary to write x'=3i and x"=i.
So I can place my points on x'Ox and I place the points A(3i,0) and B(i,0) on the horizontal axis.
All this remained very simple, and very Cartesian.
At no time did I use the Argand coordinate system (which talks about totally different things), by giving a perpendicular nature to a+ib, instead of a simply longitudinal nature in a Cartesian frame.
Imaginary number i in a Cartesian frame, and imaginary number i in an Argand frame, these are totally different things.
Here, I limit myself to talking about the use of i to find the imaginary roots of curves in a Cartesian frame.
I repeat: the Argand frame is something completely "different".

 I know exactly where to plot say, point 42+21i... Where would you place in on the plane?

If you read what I just wrote, you can very easily place your point on your Cartesian coordinate system.
It is impossible in this case (unless we open an Argand coordinate system, but that is not useful at all here) to place the point anywhere other than on the x'Ox axis. Thus, your point 42+21i only makes sense in an Argand coordinate system, and not in a Cartesian coordinate system.
But where to put it on x'Ox in the Cartesian coordinate system?
We said that the i-axis is conjunct the x'Ox axis, but inverted. This means that the abscissa 42+21i is at 42-21=21, and its ordinate at 0 (all ordinates are at y=0).
So we have your point which is at A(21,0).
If your complex was 42-21i, it would be in A(63,0) which you can also write as A(-63i,0), it is the same thing, written differently.
Be careful, I repeat, here, we are in Cartesian frames,
where i is only a logitudinal elogation carried to a fixed number, and not in Argand frames which visualize something completely different, and according to an orthogonal representation.
R.H.