x²+4x+5=0

This equation has no root, and it never will.

We can then find two roots of its mirror curve.

Let x'=-3 and x"=-1

These are not roots of this curve, but the roots of the imaginary mirror
curve.

What is this imaginary mirror curve?

It is the curve with equation y=-x²-4x-3

Let's look for its roots, and we find x'=-3 and x'=-1

These are the imaginary roots of x²-4x+5.

Or x'=-3(i) and x'=-1(i)

R.H.

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Le 22/01/2025 à 14:48, Richard Hachel a écrit :

x²+4x+5=0

This equation has no root, and it never will.

We can then find two roots of its mirror curve.

Let x'=-3 and x"=-1

These are not roots of this curve, but the roots of the imaginary mirror curve.

What is this imaginary mirror curve?

It is the curve with equation y=-x²-4x-3

Let's look for its roots, and we find x'=-3 and x'=-1

These are the imaginary roots of x²-4x+5.

Or x'=-3(i) and x'=-1(i)

R.H.

I wrote all this very badly.

The equation has only one and the same root (double) which is obviously
x=-2+i

This is the imaginary solution for y=x²+4x+5 which has no solution in R.

It is at the same time the solution for its mirror curve y=-x²-4x-3 (we
give i the values ​​-1 and 1, and we find x=-3 and x=-1.

R.H.

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Le 23/01/2025 à 11:51, Peter Fairbrother a écrit :

On 22/01/2025 13:48, Richard Hachel wrote:

x²+4x+5=0

This equation has no root, and it never will.

-3.6180339887499
-1.3819660112501

Peter Fairbrother

On calcule comme ça en Angleterre?

Dans le pays des Beatles, de Churchill et de Nigel Farage?

R.H.

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Le 23/01/2025 à 22:23, Moebius a écrit :

Am 23.01.2025 um 00:58 schrieb sobriquet:

Op 22/01/2025 om 14:48 schreef Richard Hachel:

x²+4x+5=0

This equation has no root, and it never will.

We can then find two roots of its mirror curve.

Let x'=-3 and x"=-1
What is this imaginary mirror curve?

It is the curve with equation y=-x²-4x-3

Let's look for its roots, and we find x'=-3 and x'=-1

These are the imaginary roots of x²-4x+5.

Wolfram Alpha tells us there are two roots:

https://www.wolframalpha.com/input?i=solve+x%5E2%2B4x%2B5%3D0

Wolfram Alpha must be wrong! :-P

No, here it is fine. The two imaginary roots are x'=-2-i and x"=-2+i

But that is not the question.

The question was:
Are the two imaginary roots the real reflection of the roots of the mirror curve (the two vertices
being conjunct)?

La courbe miroir étant donnée comme λ(x) = -a.x² - b.x -
[[b²/(2a)]+c] soit λ(x) =-x²-4x-3


R.H.

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Le 24/01/2025 à 14:15, FromTheRafters a écrit :

Richard Hachel wrote :

Le 23/01/2025 à 22:23, Moebius a écrit :

Am 23.01.2025 um 00:58 schrieb sobriquet:

Op 22/01/2025 om 14:48 schreef Richard Hachel:

x²+4x+5=0

This equation has no root, and it never will.

We can then find two roots of its mirror curve.

Let x'=-3 and x"=-1
What is this imaginary mirror curve?

It is the curve with equation y=-x²-4x-3

Let's look for its roots, and we find x'=-3 and x'=-1

These are the imaginary roots of x²-4x+5.

Wolfram Alpha tells us there are two roots:

https://www.wolframalpha.com/input?i=solve+x%5E2%2B4x%2B5%3D0

Wolfram Alpha must be wrong! :-P

No, here it is fine. The two imaginary roots are x'=-2-i and x"=-2+i

But that is not the question.

Shouldn't imaginary roots be on the y axis?

Ce ne serait plus résoudre les racines de x, mais les racines de y quand
y=0, ce qui est absurde.

Non, non, il s'agit de trouver les racines de x lorsque y=0.

C'est donc sur l'axe des x. Ici, il n'y en a pas. Les deux branches de la courbe partent vers le haut
du point A=(-2,1).

On trouve alors des racines imaginaires en imaginant la courbe miroir en vertical par rapport à M et en posant i=(+-)1.

R.H.

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Le 24/01/2025 à 16:48, Richard Hachel a écrit :

Le 24/01/2025 à 14:15, FromTheRafters a écrit :

Richard Hachel wrote :

Le 23/01/2025 à 22:23, Moebius a écrit :

Am 23.01.2025 um 00:58 schrieb sobriquet:

Op 22/01/2025 om 14:48 schreef Richard Hachel:

x²+4x+5=0

This equation has no root, and it never will.

We can then find two roots of its mirror curve.

Let x'=-3 and x"=-1
What is this imaginary mirror curve?

It is the curve with equation y=-x²-4x-3

Let's look for its roots, and we find x'=-3 and x'=-1

These are the imaginary roots of x²-4x+5.

Wolfram Alpha tells us there are two roots:

https://www.wolframalpha.com/input?i=solve+x%5E2%2B4x%2B5%3D0

Wolfram Alpha must be wrong! :-P

No, here it is fine. The two imaginary roots are x'=-2-i and x"=-2+i

But that is not the question.

Shouldn't imaginary roots be on the y axis?

Ce ne serait plus résoudre les racines de x, mais les racines de y quand y=0,
ce qui est absurde.

Non, non, il s'agit de trouver les racines de x lorsque y=0.

FromTheRafters was pointing out your misuse of the standard terminology.

if x,y are real numbers, then x + iy is a complex numbers. The term
"imaginary" (in French "imaginaire pur") denotes numbers of the form i*y.
They are on the "y axis" refers to the representation of C as coordinates
in the Euclidean plane.

So "-2 - i" and "-2 + i" are not "imaginary".

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Le 24/01/2025 à 17:53, Python a écrit :

Le 24/01/2025 à 16:48, Richard Hachel a écrit :

So "-2 - i" and "-2 + i" are not "imaginary".

Comme dans la théorie de la relativité, tout n'est pas toujours très
simple.

En jardinerie, les choses sont simples.

"Vous prenez trois graines de pâtisson, vous les mettez en pot vers le 15 avril, et vous repiquez quand les gelées ne sont plus à craindre en
gardant le meilleur pied".

Je sais parfaitement ce que c'est que trois, ce que c'est qu'un pâtisson,
ce que c'est qu'un pot,
ce que c'est que le mois d'avril, ce que c'est que repiquer, et ce que
c'est que les gelées.

Par contre, en science, existent souvent des concepts qui ne sont pas
simples pour tous.

Si je dis : "les vitesses apparentes doivent rester réciproques et
respecter le principe de covariance",
ou "Si les distances parcourues sont égales et mesurées en des temps
égaux, alors les temps propres seront égaux", Python se noie.

Si je dis "Il ne faut pas confondre accroissement des temps impropres, et courbe des extrémités des temps impropres sinon ça fausse le rapport
To/Tr dans les référentiels accélérés" Paul B Andersen se noie.

Peut se joindre alors le brouillard des mots. Qu'est ce qu'un hyperplan de simultanéité, qu'est ce qu'ue anisochronie universelle, qu'est ce qu'une chronotropie relative? Qu'est ce qu'un effet Doppler interne plus compréhensible que effet Doppler transversal)?

Ici, nous parlons de nombres complexes. Pour moi, ce sont des nombres imaginaires de par leur composante imaginaire multiple de i.

Z est imaginaire (sauf si b=0).

R.H.

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Le 24/01/2025 à 19:01, Moebius a écrit :

But RH told us:

"[I] call these imaginary roots, because, since they do not exist, we
must imagine them."

:-)

Well yes.

If I draw the curve x²+4x+5, I see that it has no root in reality.

I can then imagine the mirror curve adjacent to the vertex, which is
-x²-4x-3

I then have the roots x'=-3 and x"=-1

Which is the equivalent of the imaginary concept x=-2+i (with both i=-1
and i=+1).

R.H.

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Le 24/01/2025 à 19:30, Richard Hachel a écrit :

Le 24/01/2025 à 19:01, Moebius a écrit :

But RH told us:

"[I] call these imaginary roots, because, since they do not exist, we
must imagine them."

:-)

Well yes.

If I draw the curve x²+4x+5, I see that it has no root in reality.

Not "in reality": in R. It has roots in a different, bigger, set: C.

I can then imagine the mirror curve adjacent to the vertex, which is -x²-4x-3

Your "imagination" is off-topic. Another polynomial has different roots?
So what? They are not roots of x²+4x+5 anyway.

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Le 24/01/2025 à 19:01, Moebius a écrit :

Am 24.01.2025 um 17:53 schrieb Python:

Le 24/01/2025 à 16:48, Richard Hachel  a écrit :

Le 24/01/2025 à 14:15, FromTheRafters a écrit :

Richard Hachel wrote :

The two imaginary roots are x'=-2-i and x"=-2+i.

Shouldn't imaginary roots be on the y axis?

Ce ne serait plus résoudre les racines de x, mais les racines de y
quand y=0, ce qui est absurde.
Non, non, il s'agit de trouver les racines de x lorsque y=0.

FromTheRafters was pointing out your misuse of the standard terminology.

if x,y are real numbers, then x + iy is a complex numbers. The term
"imaginary" (in French "imaginaire pur") denotes numbers of the form
i*y. They are on the "y axis" refers to the representation of C as
coordinates in the Euclidean plane.

So "-2 - i" and "-2 + i" are not "imaginary".

But RH told us:

"[I] call these imaginary roots, because, since they do not exist, we
must imagine them."

:-)

He is not the sharpest knife in the drawer as you've noticed.

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Le 24/01/2025 à 19:12, Richard Hachel a écrit :

Le 24/01/2025 à 17:53, Python a écrit :

Le 24/01/2025 à 16:48, Richard Hachel a écrit :

So "-2 - i" and "-2 + i" are not "imaginary".

Comme dans la théorie de la relativité [snip idiotic babbling]

Relativity is off-topic here. Anyway you are also wrong in Relativity.

Si je dis : "les vitesses apparentes doivent rester réciproques et respecter le
principe de covariance",
ou "Si les distances parcourues sont égales et mesurées en des temps égaux,
alors les temps propres seront égaux", Python se noie.

[snip more idiotic babbling]

I've proven you wrong on both of this issues. But this is still off-topic.

Ici, nous parlons de nombres complexes. Pour moi, ce sont des nombres imaginaires de par leur composante imaginaire multiple de i.

Z est imaginaire (sauf si b=0).

This is not the usual terminology.

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Le 24/01/2025 à 20:45, Python a écrit :

Your "imagination" is off-topic. Another polynomial has different roots? So what? They are not roots of x²+4x+5 anyway.

I proposed the curve y=x²+4x+5 and we showed that it had no real roots.
I then proposed to study the mirror curve centered on its vertex, and to
look at where the two real roots were.
I said that this curve necessarily had the aspect of f'(x)=-x²-4x+c' with c'=c-(b²/2a)
Julien also says the same thing.

And so f'(x)=-x²-4x-3

Two roots in R: x'=-3 and x"=-1

The two roots of the first curve were x'=-2-i and x"=-2+i

It therefore seems that the imaginary roots of a curve without real roots
have a resemblance to the real roots of the mirror curve.

R.H.

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Le 24/01/2025 à 20:56, Python a écrit :

Le 24/01/2025 à 19:12, Richard Hachel a écrit :

Z est imaginaire (sauf si b=0).

This is not the usual terminology.

C'est ce que je dis.

J'ai souvent remarqué que les scientifiques, ne sachant pas toujours de façon claire ce dont ils parlent,
ont des terminologies foireuses ou douteuses.

Par exemple, ils parlent d'effet Doppler transversal, ce qui est une contradiction dans les termes.

Mon immense intelligence me fait comprendre qu'ils veulent dire par là
qu'il s'agit d'un effet relativiste spécial dans le sens où lorsqu'on
observe un corps se déplaçant de façon transversale ,ous avons là un
effet Doppler absolument innocent d'un quelconque effet longitudinal.

C'est cela qu'ils veulent dire.

Mais c'est très mal dit.

Je préfère parler d'effet Doppler interne. C'est beaucoup plus juste et
plus indicatif.

C'est pareil pour les nombres imaginaire. Z est un nombre imaginaire parce qu'il est composé d'une partie imaginaire

R.H.

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Le 02/02/2025 à 23:55, "Chris M. Thomasson" a écrit :

On 1/23/2025 2:51 AM, Peter Fairbrother wrote:

On 22/01/2025 13:48, Richard Hachel wrote:

x²+4x+5=0

This equation has no root, and it never will.

-3.6180339887499
-1.3819660112501

-2-1i
-2+1i

The two roots have non-zero imaginary parts, so they are not reals.

(-2-1i)²+4(-2-1i)+5 = 0
(-2+1i)²+4(-2+1i)+5 = 0

:^)

This is obviously very true.
But it poses a small problem for me that I can't quite explain.
I'll think about that.

R.H.

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