Path: ...!weretis.net!feeder8.news.weretis.net!pasdenom.info!from-devjntp
Message-ID: <bwfbOvWBAxirGh8NBCIO1qXfaj0@jntp>
JNTP-Route: nemoweb.net
JNTP-DataType: Article
Subject: Re: New equation
References: <CJRYb90R4pKx6LHEBcCdcCA4y30@jntp> <d2lprjtieerl4rjrqs9s878j0d03jm645q@4ax.com> <MimQ6gJUHMh7Qd3O8h9HuwqFhUg@jntp>
 <15vqrj5iscra2hlaiukp60qo0mkiquvai8@4ax.com> <esk4yUvD7xtbFRMwGfaB_cgVGwg@jntp> <vpld4o$26f02$1@dont-email.me>
 <ZuBkOZlN3Tt29PxIPLwN6BSCDt0@jntp> <vpli8o$27a25$2@dont-email.me> <paGqSVDKFAulO0cdlEqGbkSJSws@jntp>
 <vpllt6$27sv3$2@dont-email.me>
Newsgroups: sci.math
JNTP-HashClient: Wveoa-qOfu4VMZIfCn6_s5UrMRE
JNTP-ThreadID: 18KZ-4JA0HrCG_MvALF7ypvYK-g
JNTP-Uri: https://www.nemoweb.net/?DataID=bwfbOvWBAxirGh8NBCIO1qXfaj0@jntp
User-Agent: Nemo/1.0
JNTP-OriginServer: nemoweb.net
Date: Wed, 26 Feb 25 00:23:52 +0000
Organization: Nemoweb
JNTP-Browser: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/133.0.0.0 Safari/537.36
Injection-Info: nemoweb.net; posting-host="0622b338f00df6c7e122ad5f6ee90645acf995aa"; logging-data="2025-02-26T00:23:52Z/9222254"; posting-account="4@nemoweb.net"; mail-complaints-to="julien.arlandis@gmail.com"
JNTP-ProtocolVersion: 0.21.1
JNTP-Server: PhpNemoServer/0.94.5
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
X-JNTP-JsonNewsGateway: 0.96
From: Richard Hachel <r.hachel@tiscali.fr>
Bytes: 5672
Lines: 77

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.