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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: New equation
Date: Tue, 25 Feb 2025 16:05:58 -0800
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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?