Equation complexe

x^4=-81

What is x?

R.H.

Jim Burns

2025-02-25 17:21:23 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
⎛ ...
⎜ every non-constant single-variable polynomial
⎜ with complex coefficients has at least one complex root.
⎜
⎜ The theorem is also stated as follows:
⎜ every non-zero, single-variable, degree n polynomial
⎜ with complex coefficients has,
⎝ counted with multiplicity, exactly n complex roots.

https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Proofs
Real-analytic proofs
Complex-analytic proofs
Topological proofs
Algebraic proofs
By induction
From Galois theory
Geometric proofs

Richard Hachel

2025-02-25 20:07:02 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Oui, c'est ce que dis aussi l'Intelligence artificielle, mais sans trop
expliquer pourquoi.

Personally, I propose only one root, but it is not in conformity with
what is said about "complex numbers".
I remind you that I do not admit the definition i²=-1, which, in itself,
is not false, but so narrow that I do not understand its semantic
interest.
For me, the definition must be extended to all powers of x such that
i^x=-1.
In short, it is necessary to go through a constancy of i^x=-1, like 1^x=1.
This defined, we easily have x^4=-81 if x=3i.
The other four roots being incorrect (in the proposed system).

R.H.

Python

2025-02-25 20:38:27 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Oui, c'est ce que dis aussi l'Intelligence artificielle, mais sans trop
expliquer pourquoi.
Personally, I propose only one root, but it is not in conformity with what is
said about "complex numbers".

So it is not about complex as it is not only said but proven from the
definition of complex numbers.

Post by Richard Hachel
I remind you that I do not admit the definition i²=-1,

It is not the definition, it is a consequence of a definition that I've
shown you numerous times. You are a liar.

Post by Richard Hachel
which, in itself, is not false, but so narrow that I do not understand its
semantic interest
For me, the definition must be extended to all powers of x such that i^x=-1.

Which leads to immediate inconsistencies. This property is contradictory.

Post by Richard Hachel
In short, it is necessary to go through a constancy of i^x=-1, like 1^x=1.
This defined, we easily have x^4=-81 if x=3i.
The other four roots being incorrect (in the proposed system).

This is utter bullshit. As are all your your fantasies. Build on top of
your ignorance, hypocrisies, stupidity and arrogance.

Richard Hachel

2025-02-25 21:47:14 UTC

Post by Richard Hachel
For me, the definition must be extended to all powers of x such that i^x=-1.

This property is contradictory.

Absolutely not.

Tu n'as toujours pas compris (comme tu n'as toujours pas compris toutes
les équations que j'ai données
pour bonifier et rendre plus compréhensible la relativité restreinte).

This is not inserting a contradictory property, contradictory properties,
in all my special relativity, I have knocked them out.
It is the complete opposite of my personality to propose contradictory
properties.
On the contrary, it is a fantastic basic principle that is affirmed.
We extend the property of the imaginary number i to all powers of i. The
definition of i is then clearly named, and given.
The imaginary number i is a very useful imaginary number to perform many
operations and whose definition is the invariance of its nature whatever
the power that we give it.
i^x=-1 whatever x. We see that i is the antithesis of 1.
It then becomes obvious that i^(1/2)=-1 and that i²=-1 as many say. But
this is not enough if they believe that i²*i²=1.
Here, there is a bias, we use a real operation on an imaginary structure.
We must set i²*i²=i^4=-1.
This is NOT contradictory. It is an intrinsic part of my (Nazi) ideology.

R.H.

Python

2025-02-25 21:51:41 UTC

Post by Richard Hachel
For me, the definition must be extended to all powers of x such that i^x=-1.

This property is contradictory.

It is actually a perfectly accurate definition of your personality. In
addition to fatuity, ignorance and dishonesty.

Post by Richard Hachel
On the contrary, it is a fantastic basic principle that is affirmed.
We extend the property of the imaginary number i to all powers of i. The
definition of i is then clearly named, and given.
The imaginary number i is a very useful imaginary number to perform many
operations and whose definition is the invariance of its nature whatever the power
that we give it.
i^x=-1 whatever x. We see that i is the antithesis of 1.
It then becomes obvious that i^(1/2)=-1 and that i²=-1 as many say. But this is
not enough if they believe that i²*i²=1.
Here, there is a bias, we use a real operation on an imaginary structure. We
must set i²*i²=i^4=-1.
This is NOT contradictory. It is an intrinsic part of my (Nazi) ideology.
R.H.

guido wugi

2025-02-25 22:05:14 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Contradictory, thus impossible.
If i^2=-1 then (i^2)^2 = (-1)^2 = 1 = i^4

--
guido wugi

Richard Hachel

2025-02-25 22:15:08 UTC

Post by guido wugi

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Contradictory, thus impossible.
If i^2=-1 then (i^2)^2 = (-1)^2 = 1 = i^4

Absolutely not...

Pfffff...

You are using real powers on imaginary numbers.

You don't seem to understand what I'm saying.

I am stating the precise, formal, and definitive definition that in this
imaginary system, the fundamental law is i^x=-1.

A fundamental law is not negotiable.

Now, you start to negotiate, like Python, and to tell me while crying:
i^4=1.

This is not serious behavior.

R.H.

Python

2025-02-25 22:18:29 UTC

Post by guido wugi

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2,
(-1-𝑖)⋅3/√2}

Contradictory, thus impossible.
If i^2=-1 then (i^2)^2 = (-1)^2 = 1 = i^4

Absolutely not...
Pfffff...
You are using real powers on imaginary numbers.
You don't seem to understand what I'm saying.
I am stating the precise, formal, and definitive definition that in this
imaginary system, the fundamental law is i^x=-1.
A fundamental law is not negotiable.
Now, you start to negotiate, like Python, and to tell me while crying: i^4=1.
This is not serious behavior.

Yours is not.

You are contradicting this :

If a = b then a*a = b*b

Python

2025-02-25 22:29:48 UTC

Post by guido wugi

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2,
(-1-𝑖)⋅3/√2}

Contradictory, thus impossible.
If i^2=-1 then (i^2)^2 = (-1)^2 = 1 = i^4

Absolutely not...
Pfffff...
You are using real powers on imaginary numbers.
You don't seem to understand what I'm saying.
I am stating the precise, formal, and definitive definition that in this
imaginary system, the fundamental law is i^x=-1.
A fundamental law is not negotiable.
Now, you start to negotiate, like Python, and to tell me while crying: i^4=1.
This is not serious behavior.

Yours is not.
If a = b then a*a = b*b

What is impressive, pathologically, is that you do so, and couldn't spot
how silly it is, because of your self-indulgence.

Jim Burns

2025-02-26 22:59:09 UTC

Post by Richard Hachel
A fundamental law is not negotiable.

https://plato.stanford.edu/entries/definitions/
⎛ ...
⎜ Stipulative definitions are epistemologically special.
⎜ They yield judgments with epistemological characteristics
⎜ that are puzzling elsewhere.
⎜ If one stipulatively defines a “raimex” as, say,
⎜ a rational, imaginative, experiencing being
⎜ then the judgment “raimexes are rational” is
⎝ assured of being necessary, certain, and a priori.

Definitions are not negotiable.

sobriquet

2025-02-26 04:12:44 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Oui, c'est ce que dis aussi l'Intelligence artificielle, mais sans trop
expliquer pourquoi.[..]

You have to understand that if we consider this equivalent case of
g(x) = x^4 + 9/4 and we visualize the complex roots as follows:

https://www.desmos.com/calculator/4mhohrcxlg

We're superimposing a real slice of the graph in the Cartesian plane on
top of Complex plane where we visualize the complex roots of the polynomial.

In order to understand why these are the roots of the polynomial g(x)
in the complex numbers, we have to expand our view to 3D:

https://www.desmos.com/3d/ykhhcpb3xz

Where we can visualize that polynomial as a function that maps complex
numbers to complex numbers as a combination of two surfaces in 3D
(one red surface for the real component of the output of the function
and an orange surface for the imaginary component of the output of the
function).
We can press the icons in front of the corresponding items in the list
on the left to temporarily hide those surfaces to identify the locations
in the complex plane where both components are equal to zero to see that
those are exactly the complex roots of the polynomial g(x) in the
complex number system.

sobriquet

2025-02-26 12:11:10 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Oui, c'est ce que dis aussi l'Intelligence artificielle, mais sans
trop expliquer pourquoi.[..]

You have to understand that if we consider this equivalent case of
https://www.desmos.com/calculator/4mhohrcxlg
We're superimposing a real slice of the graph in the Cartesian plane on
top of Complex plane where we visualize the complex roots of the polynomial.

Here we can visualize the real slice of the graph in 3D to understand
the relationship between the graph of the function in the real numbers
versus the graph of the roots of the function in the complex numbers:

https://www.desmos.com/3d/0l8ota7qha

Moebius

2025-02-26 12:31:33 UTC

Post by Richard Hachel
x^4=-81
What is x?

@RH:

If x ∈ IR, then x^2 ∈ IR and x^2 >= 0, and hence x^4 = (x^2)^2 >= 0.

So x can't be in IR.

Now let x = ir with r ∈ IR (and i, the imaginary unit, ∈ C).

Then x^2 = i^2 * r^2 = -1 * r^2 ∈ IR. Hence x^4 = (x^2)^2 >= 0.

Hence neither x ∈ IR nor x = ir with r ∈ IR will work.

Actually, any x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2,
(-1-𝑖)⋅3/√2} will work.

Hope this helps.

Richard Hachel

2025-02-26 18:12:21 UTC

Post by Moebius
If x ∈ IR, then x^2 ∈ IR and x^2 >= 0, and hence x^4 = (x^2)^2 >= 0.
So x can't be in IR.
Now let x = ir with r ∈ IR (and i, the imaginary unit, ∈ C).
Then x^2 = i^2 * r^2 = -1 * r^2 ∈ IR. Hence x^4 = (x^2)^2 >= 0.
Hence neither x ∈ IR nor x = ir with r ∈ IR will work.
Actually, any x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2,
(-1-𝑖)⋅3/√2} will work.
Hope this helps.

Thank you.
But we are not talking about the same thing.
Mathematicians talk about an imaginary unit i.
I also talk about an imaginary unit i.
Mathematicians propose to replace in a negative discriminant a
multiplication by 1, by a multiplication by -i², since i²=-1 and 1=-i².
I propose strictly the same thing, and, obviously, I obtain the same
results as them.
That is not the problem.
Simply, mathematicians do not explain WHY their imaginary unit i is worth
-1 when we square it. They make dictates. Useful dictates, interesting
dictates, but dictates nonetheless.
They don't seem to realize what this imaginary unit is that they use, and
only see it under a distant i²=-1 that is very limited and that explains
nothing, and that plunges them into error as soon as they use anything
other than simple quadratic equations.
If we ask them to solve a degree 4 equation without real roots, for
example, everything will collapse in horror, and they will come up with
absolutely anything.
Let's set f(x)=x^4+(9/4) or f(x)=x^4+x²+2 and the complex roots that they
give are so ridiculous that they make you laugh.
This shows that they have not understood what i was and how to correctly
handle this unit in basic analytics. I am not talking about the Argand
frame, that is something else. I am talking about the role of i in the
Cartesian frames of reference. The Argand frame is "something else" where
the modulus of a complex is defined as the square root of the product of
its two conjugate complexes.

R.H.

Python

2025-02-26 18:23:42 UTC

Thank you.
But we are not talking about the same thing.
Mathematicians talk about an imaginary unit i.

The word "imaginary" is a historic relief. We kept it but there is nothing
"imaginary" in the usual sense when it comes to i or C (no more than any
other consistently defined object such as 41, 1, 0 or -1).

Did you know that at some point in history some mathematicians rejected
negative numbers?

What do you know about modern (i.e. post XIXe century) definition of
natural numbers, relative numbers, rationals, "real" numbers, complex
numbers (and more) ?

NOTHING. You are pretending to teach without any knowledge of anything.
Also without honesty.

Post by Richard Hachel
I also talk about an imaginary unit i.
Mathematicians propose to replace in a negative discriminant a multiplication by
1, by a multiplication by -i², since i²=-1 and 1=-i².
I propose strictly the same thing, and, obviously, I obtain the same results as
them.

Obviously not.

Post by Richard Hachel
That is not the problem.
Simply, mathematicians do not explain WHY their imaginary unit i is worth -1
when we square it. They make dictates. Useful dictates, interesting dictates, but
dictates nonetheless.

You are lying again. This has been said to you in fr.sci.maths.

We can explain why i^2 = -1 from this very precise and clear definition:

i is the equivalence class of the polynomial X in the quotient ring
[actually a field] R[X]/(X^1+1).

From that you can deduce that i^2 = -1.

If you cannot get it, read my course (or others) and/or ask instead of
fantasizing on your ill-defined contradictory personal "ideas".

Post by Richard Hachel
[snip more nonsense and lies]

You are an insufferable crank, "Dr. Hachel" (aka former M.D. Richard
Lengrand).

Moebius

2025-02-26 18:34:11 UTC

Post by Python
i is the equivalence class of the polynomial X in the quotient ring
[actually a field] R[X]/(X^1+1).
From that you can deduce that i^2 = -1.

Same with C := {(x, y) : x,y e IR}, i := (0, 1) and properly defined
addition and multiplication on C.

Then we can PROVE that î * i = (-1, 0) aka -1.

No big deal.

Python

2025-02-26 18:38:39 UTC

Post by Python
i is the equivalence class of the polynomial X in the quotient ring
[actually a field] R[X]/(X^1+1).
From that you can deduce that i^2 = -1.

Same with C := {(x, y) : x,y e IR}, i := (0, 1) and properly defined
addition and multiplication on C.
Then we can PROVE that î * i = (-1, 0) aka -1.
No big deal.

Sure. There are numerous (equivalent) definition. 2x2 matrices etc.

Everything is a big deal for an hypocrite crank like Hachel/Lengrand.

Richard Hachel

2025-02-26 18:50:27 UTC

i is the equivalence class of the polynomial X in the quotient ring [actually a
field] R[X]/(X^1+1).

Tu as fait une faute de frappe.

From that you can deduce that i^2 = -1.

Yes.

Mais seulement cela.

Le reste est complétement foireux.

Tu n'as pas le droit d'imposer i²*i²=i^4=1

C'est la même chose que de dire si x²=1 alors racine unique x=-1. C'est
débile.

Tu veux finir aussi bouffon que efji?

R.H.

Python

2025-02-26 18:54:20 UTC

i is the equivalence class of the polynomial X in the quotient ring [actually a
field] R[X]/(X^1+1).

Tu as fait une faute de frappe.

I made no typo up there.

From that you can deduce that i^2 = -1.

Yes.
Mais seulement cela.

Far more than that actually. How could you know? You didn't check.

Python

2025-02-26 18:56:54 UTC

i is the equivalence class of the polynomial X in the quotient ring [actually a
field] R[X]/(X^1+1).
From that you can deduce that i^2 = -1.

Yes.

Well... Could you show it? I can.

Richard Hachel

2025-02-26 19:13:50 UTC

i is the equivalence class of the polynomial X in the quotient ring [actually a
field] R[X]/(X^1+1).
From that you can deduce that i^2 = -1.

Yes.

Well... Could you show it? I can.

Can you show what i is the equivalence class of the polynomial X in the
quotient ring [actually a field] R[X]/(X^4+1)?

R.H.

Python

2025-02-26 19:18:28 UTC

i is the equivalence class of the polynomial X in the quotient ring [actually a
field] R[X]/(X^1+1).
From that you can deduce that i^2 = -1.

Yes.

Well... Could you show it? I can.

Can you show what i is the equivalence class of the polynomial X in the quotient
ring [actually a field] R[X]/(X^4+1)?

This is a *definition* and definitely NOT you lie about "posing than i^2 =
-1 as a definition".

The point is to show that this definition is consistent (it is) and to
show that, then, i^2 = -1.

Again: you wrote "Yes". So can you show it? Or will you chicken out as
usual?

Python

2025-02-26 19:23:56 UTC

i is the equivalence class of the polynomial X in the quotient ring [actually a
field] R[X]/(X^1+1).
From that you can deduce that i^2 = -1.

Yes.

Well... Could you show it? I can.

Can you show what i is the equivalence class of the polynomial X in the quotient
ring [actually a field] R[X]/(X^4+1)?

Fucking dishonest kook!!! You changed X^2 into X^4!!!

Did you even understand a single word of your own posts? Clearly not.

So, again:

You answered "Yes" to: "i is the equivalence class of the polynomial X in
the quotient ring [actually a field] R[X]/(X^1+1)."

Can you show it to start with? I can.

Richard Hachel

2025-02-26 19:43:37 UTC

Can you show what i is the equivalence class of the polynomial X in the quotient
ring [actually a field] R[X]/(X^4+1)?

You changed X^2 into X^4!!!

Yes.

Can you?

R.H.

Python

2025-02-26 19:45:09 UTC

Can you show what i is the equivalence class of the polynomial X in the quotient
ring [actually a field] R[X]/(X^4+1)?

You changed X^2 into X^4!!!

Yes.
Can you?
R.H.

You'll get once, Lengrand, the correction you deserve, you know? In
person.

Moebius

2025-02-27 07:16:12 UTC

x^4 = -81
What is x?

If x ∈ IR, then x^2 ∈ IR and x^2 >= 0, and hence x^4 = (x^2)^2 >= 0.
So x can't be in IR.
Now let x = ir with r ∈ IR (and i, the imaginary unit, ∈ C).
Then x^2 = i^2 * r^2 = -1 * r^2 ∈ IR. Hence x^4 = (x^2)^2 >= 0.
Hence neither x ∈ IR nor x = ir with r ∈ IR will work.
Actually, any x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}
will work.

After all, playing by the rules we get:

((+1+𝑖)⋅3/√2)^4 = -81
((+1-𝑖)⋅3/√2)^4 = -81
((-1+𝑖)⋅3/√2)^4 = -81
((-1-𝑖)⋅3/√2)^4 = -81

Moreover: It can be shown that there are no other x e C such that x^4 =
-81. (See: Fundamental theorem of algebra)

Post by Moebius
Hope this helps.

Richard Hachel

2025-02-27 16:20:34 UTC

Post by Moebius
((+1+𝑖)⋅3/√2)^4 = -81
((+1-𝑖)⋅3/√2)^4 = -81
((-1+𝑖)⋅3/√2)^4 = -81
((-1-𝑖)⋅3/√2)^4 = -81

((+1+𝑖)⋅3/√2)^4 = -81
((+1+𝑖)^4(3/√2)^4 = -81
(1+2i+i²)(1+2i+i²)(3/√2)^4=-81
i²=-1
(2i²)(81/4)=(i²)(81/2)=-81
i²=-1
-40.5=-81

C'est absurde.

Etc....

Non, non, non, non...

(3i²)^4=-81
i^4=-1 (Hachel copyrights)
3^4=81

R.H.

sobriquet

2025-02-27 20:48:27 UTC

Post by Moebius
((+1+𝑖)⋅3/√2)^4 = -81
((+1-𝑖)⋅3/√2)^4 = -81
((-1+𝑖)⋅3/√2)^4 = -81
((-1-𝑖)⋅3/√2)^4 = -81

((+1+𝑖)⋅3/√2)^4 = -81
((+1+𝑖)^4(3/√2)^4 = -81
(1+2i+i²)(1+2i+i²)(3/√2)^4=-81
i²=-1
(2i²)(81/4)=(i²)(81/2)=-81

Here you go wrong.. it should say:
(2i)²(81/4)=(i²)81=-81

Post by Moebius
i²=-1
-40.5=-81
C'est absurde.
Etc....
Non, non, non, non...
(3i²)^4=-81
i^4=-1 (Hachel copyrights)
3^4=81
R.H.

Richard Hachel

2025-02-26 17:53:01 UTC

Post by sobriquet
g(x) = x^4 + 9/4
https://www.desmos.com/calculator/4mhohrcxlg

? ? ?

h(x)=-x^4+(9/4)

Racines réelles h(x)----> x'=-sqrt(3/2) et x"=sqrt(3/2)

Complexes roots of g(x) : x'=sqrt(3/2)i and x"=-sqrt(3/2)i

R.H.

sobriquet

2025-02-26 19:15:11 UTC

Post by sobriquet
g(x) = x^4 + 9/4
https://www.desmos.com/calculator/4mhohrcxlg

? ? ?
h(x)=-x^4+(9/4)
Racines réelles h(x)----> x'=-sqrt(3/2) et x"=sqrt(3/2)
Complexes roots of g(x) : x'=sqrt(3/2)i and x"=-sqrt(3/2)i
R.H.

You started the thread with the equation x^4=-81, but
I've picked the equation x^4=-9/4 instead, since that means
the graph is closer to the complex roots when I visualize it in
desmos.

The polynomial g(x)= x^4 + 9/4 is equivalent to the equation x^4=-9/4.

Richard Hachel

2025-02-26 19:17:37 UTC

Post by sobriquet
g(x) = x^4 + 9/4
https://www.desmos.com/calculator/4mhohrcxlg

? ? ?
h(x)=-x^4+(9/4)
Racines réelles h(x)----> x'=-sqrt(3/2) et x"=sqrt(3/2)
Complexes roots of g(x) : x'=sqrt(3/2)i and x"=-sqrt(3/2)i
R.H.

You started the thread with the equation x^4=-81, but
I've picked the equation x^4=-9/4 instead, since that means
the graph is closer to the complex roots when I visualize it in
desmos.
The polynomial g(x)= x^4 + 9/4 is equivalent to the equation x^4=-9/4.

Absolutely.

And I gived the correct roots.

R.H.

Jim Burns

2025-02-26 17:57:05 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Oui, c'est ce que dis aussi l'Intelligence artificielle,
mais sans trop expliquer pourquoi.

(x-(+1+𝑖)⋅3/√2)⋅(x-(+1-𝑖)⋅3/√2)⋅(x-(-1+𝑖)⋅3/√2)⋅(x-(-1-𝑖)⋅3/√2) =
x⁴+81

In real and in complex numbers,
if a⋅b⋅c⋅d = 0
then one of a,b,c,d is 0
otherwise, a⋅b⋅c⋅d ≠ 0

If x⁴+81 = 0
and a⋅b⋅c⋅d = x⁴+81
then one of
a = x-(+1+𝑖)⋅3/√2 = 0
b = x-(+1-𝑖)⋅3/√2 = 0
c = x-(-1+𝑖)⋅3/√2 = 0
d = x-(-1-𝑖)⋅3/√2 = 0
is true.

Hence,
x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Post by Richard Hachel
Personally, I propose only one root,
but it is not in conformity with
what is said about "complex numbers".

Therefore, despite appearances,
your question is not "What is x?"
and it is not "what is a complex number?"

Your question is
"How can it be possible
for one person to speak to another
and be understood?"

The possibility of understanding is
greatly facilitated where
that which a speaker means by a word and
that which a listener thinks they mean by it
are in conformity.

This is not a deep philosophical point,
This is similar to noting that
phone calls with non.operating phones
are completely unsatisfactory.
A point doesn't need to be deep to be true.

Post by Richard Hachel
I remind you that
I do not admit the definition i²=-1,

Your question assumes a certain common background,
because that's how language works.
If your question was "How do I get to the post office?"
but you didn't admit the usual definitions of
'left', 'right', and so on,
odds are you don't get to the post office,
but that would have nothing to do with
the directions you were given.

Post by Richard Hachel
which, in itself, is not false, but so narrow
that I do not understand its semantic interest.

I think you are asking why 𝐢²=-1
and why not = something else.

Most of the answer is that
we want a 2.dimensional field which extends
the 1.dimensional field of the real numbers.
That is to say, we want 2.dimensional
addition '+' and multiplication '⋅'
which satisfy the same laws which
our 1.dimensional '+' and '⋅' satisfy:
⎛ associativity and commutativity for both,
⎜ identities 𝟎 𝟏, inverses -𝐱 𝐱⁻¹ except 𝟎⁻¹,
⎝ distributivity of '.' over '+'

We have what we want if
we have a vector 𝐯 not on the real axis such that,
for this 2.dimensional multiplication '⋅'
𝟏⋅𝟏 = 𝟏
𝟏⋅𝐯 = 𝐯
𝐯⋅𝟏 = 𝐯
𝐯⋅𝐯 = -α𝟏-2β𝐯
such that α > β²

Pick 𝐯 ∈ ℝ×(ℝ\{0}), β, α > β²
Define
𝐯⋅𝐯 = -α𝟏-2β𝐯
(a𝟏+b𝐯)⋅(c𝟏+d𝐯) = ac𝟏+(ad+bd)𝐯+bd(𝐯⋅𝐯)

We have what we want,
a 2.dimensional field extending ℝ

However,
suppose 𝐯 ≠ ⟨0,1⟩ and 𝐯⋅𝐯 ≠ -𝟏
Then 𝐯 ≠ 𝐢

But 𝐢 still exists,//////////////////
determined by choices 𝐯, β, α
𝐢 = ±(𝐯+β𝟏)/(α-β²)¹ᐟ² (either works)
𝐯 = ±(α-β²)¹ᐟ²𝐢-β𝟏

Each a𝟏+b𝐯 has a corresponding a′𝟏+b′𝐢
and vice versa.
𝟏⋅𝟏 = 𝟏
𝟏⋅𝐢 = 𝐢
𝐢⋅𝟏 = 𝐢
𝐢⋅𝐢 = -𝟏
(a𝟏+b𝐢)⋅(c𝟏+d𝐢) = (ac-bd)𝟏+(ad+bd)𝐢
in the usual way.

Post by Richard Hachel
For me, the definition must be
extended to all powers of x such that
i^x=-1.

For me, the definition of 'left' must be
extended to all directions.
Wish me luck!

Post by Richard Hachel
The other four roots being incorrect
(in the proposed system).

The proposed system do not have
2.dimensional '+' and '⋅' satisfying:
⎛ associativity and commutativity for both,
⎜ identities 𝟎 𝟏, inverses -𝐱 𝐱⁻¹ except 𝟎⁻¹,
⎝ distributivity of '.' over '+'

Richard Hachel

2025-02-26 18:43:37 UTC

Post by Richard Hachel
x^4=-81
What is x?

x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Oui, c'est ce que dis aussi l'Intelligence artificielle,
mais sans trop expliquer pourquoi.

(x-(+1+𝑖)⋅3/√2)⋅(x-(+1-𝑖)⋅3/√2)⋅(x-(-1+𝑖)⋅3/√2)⋅(x-(-1-𝑖)⋅3/√2)
=
x⁴+81
In real and in complex numbers,
if a⋅b⋅c⋅d = 0
then one of a,b,c,d is 0
otherwise, a⋅b⋅c⋅d ≠ 0

You are absolutely right.
I will allow myself to frame it in my newsreader.

In real and in complex numbers, if a⋅b⋅c⋅d = 0
then one of a,b,c,d is 0 ; otherwise, a⋅b⋅c⋅d ≠ 0

Post by Jim Burns
If x⁴+81 = 0
and a⋅b⋅c⋅d = x⁴+81
then one of
a = x-(+1+𝑖)⋅3/√2 = 0
b = x-(+1-𝑖)⋅3/√2 = 0
c = x-(-1+𝑖)⋅3/√2 = 0
d = x-(-1-𝑖)⋅3/√2 = 0
is true.

No.

Here we are in "the problem of compensated error".
There is a basic error in the calculation of the roots, and we make the
same error going back to the base.

But the error will quickly be found if we use the unit i as it must be
used in this case.

For example :
x⁴+81 = 0
[(+1+𝑖)⋅3/√2]^4+81=0 ---> 81=0
[(+1-𝑖)⋅3/√2]^4+81=0 ---> (6/√2)^4+81=0 ---> 405=0
[(-1+𝑖)⋅3/√2]^4+81=0 ---> (-6/√2)^4+81=0 ---> 405=0
[(-1-𝑖)⋅3/√2]^4+81=0 ----> 81=0

It's not serious.

If you take x⁴+81 = 0 with i⁴=-1 (Hachel's postulat) :

x⁴=-81
x⁴=-81*(1)
x⁴=-81*(-i⁴)=81.i⁴
x=3i

R.H.

Python

2025-02-26 18:48:12 UTC

Le 26/02/2025 à 19:43, Richard Hachel a écrit :
..

Post by Richard Hachel
You are absolutely right.
...
In real and in complex numbers, if a⋅b⋅c⋅d = 0 then one of a,b,c,d is 0 ;
otherwise, a⋅b⋅c⋅d ≠ 0

In your previous "system", this was false: there were divisors of 0 i.e.
non-zero items z1, z2 such as z1*z2 = 0. You didn't even notice it first,
I pointed it out it on fr.sci.maths.

You are a damn hypocrite Lengrand.

Richard Hachel

2025-02-26 19:10:47 UTC

Post by Python
In your previous "system", this was false: there were divisors of 0 i.e.
non-zero items z1, z2 such as z1*z2 = 0. You didn't even notice it first, I
pointed it out it on fr.sci.maths.

This is entirely correct.
And you were absolutely right to point this out.
In the Hachel system, we have Z=z1+z2=(a+a')+i(b+b').
Which is consistent with the traditional mathematical system.
But we do not have Z=z1.z2=(aa'-bb')+i(ab'+a'b)
but Z=z1.z2=(aa'+bb')+i(ab'+a'b).
So far no problem, it may be wrong (which I don't think),
but it is consistent.
Now, the inverse operation which is the quotient, will obviously also be
upset, and, we have, in the Hachel formula a divisor a'+ib' which will
induce in the denominator a'²-b'².
This is what Jean-Paul Messager noticed. What does this mean?
This means that, for example, you cannot divide a complex by another
complex of type a'+ib' if a'=b'.
On the surface, it may sound funny to say that you cannot divide a complex
by 5+5i for example.
But it makes sense.
This amounts to dividing by zero without us realizing it.

R.H.

Python

2025-02-26 19:16:19 UTC

Consistent, but without relation with complex numbers and useless.

Post by Richard Hachel
Now, the inverse operation which is the quotient, will obviously also be upset,
and, we have, in the Hachel formula a divisor a'+ib' which will induce in the
denominator a'²-b'².
This is what Jean-Paul Messager noticed. What does this mean?
This means that, for example, you cannot divide a complex by another complex of
type a'+ib' if a'=b'.
On the surface, it may sound funny to say that you cannot divide a complex by
5+5i for example.
But it makes sense.
This amounts to dividing by zero without us realizing it.
R.H.

At least your previous system were consistent, it was not at all
describing complex numbers, but it was consistent as a ring (not a field,
so irrelevant when it comes to what complex numbers allows).

Your new "system" with i^2 = -1 and i^4 = -1 is not even consistent. It
can be shown contradictory in one line. [note that this systems is NOT the
previous one you proposed at all]

Meanwhile you still fail to understand what complex numbers are, why they
are consistent, useful and used for: you are drowning yourself in
stupidity, ignorance and arrogance. You persist in lies and hypocrisy. In
other words: the story of your life.

Richard Hachel

2025-02-26 19:32:05 UTC

Post by Python
Your new "system" with i^2 = -1 and i^4 = -1 is not even consistent.

It is the system i²*i²=i^4=1 that does not seem coherent to me.

Of course, in the real world, we have for example (x²)² positive.

But here, we have introduced an imaginary unit, such as i^x=-1.

This is a definition that will not simply apply to the roots of quadratic
functions, but to ALL functions.

Including for example f(x)=sqrt(x)+2 or g(x)=(x²)²+x²+2 which are other
types of functions than a simple quadratic.

I can easily get the complex roots out of it (there are no real roots).

The roots given by mathematicians are not correct, because they are based
on an Argand frame, with a separation of a and ib into two orthogonal
entities, which does not exist in Cartesian frames where complex roots are
on x'Ox like the others, and are the object of simple additions such as
x'=5+3i=2, or x"=8-2i=10 (i=-1).

R.H.

Python

2025-02-26 19:38:27 UTC

Post by Python
Your new "system" with i^2 = -1 and i^4 = -1 is not even consistent.

It is the system i²*i²=i^4=1 that does not seem coherent to me.

With no sensible reason. Your "feelings" are irrelevant.

Post by Richard Hachel
Of course, in the real world, we have for example (x²)² positive.
But here, we have introduced an imaginary unit, such as i^x=-1.

You are again using "real" and "imaginary" as words in the common
language. This is the root of your confusion. I've explained that to you
numerous time.

There is nothing "real" or "imaginary" in 0, -1, 42 or i. They are now
technical terms with proper definition. THIS IS NOT A FAIRY TAIL!!!
Idiot!!!

Post by Richard Hachel
This is a definition that will not simply apply to the roots of quadratic
functions, but to ALL functions.
Including for example f(x)=sqrt(x)+2 or g(x)=(x²)²+x²+2 which are other types
of functions than a simple quadratic.
I can easily get the complex roots out of it (there are no real roots).

You get no roots at all. Your claims are 100% nonsense.

Post by Richard Hachel
The roots given by mathematicians are not correct, because they are based on an
Argand frame,

This is FALSE. Argand's plane is not involved in the proof that they are
roots. You are making up stuff as usual.

There no point in discussing with an ignorant, arrogant, stupid and
dishonest person like you, Lengrand. I'll continue to point out your
contradictions and lies anyway.

Richard Hachel

2025-02-26 19:40:23 UTC

I'll continue to point out your contradictions and lies anyway.

Good luck. :))

R.H.

Moebius

2025-02-27 07:05:38 UTC

Your new "system" with i² = -1 and i^4 = -1 is not even consistent.

Indeed, especially if i^4 = i² * i².

After all, if i² = -1, then i^4 = i² * i² = (-1) * (-1) = 1.

And 1 =/= -1.

It is the system i² * i² = i^4 = 1 that does not seem coherent to me.

Strange.

After all, if i² = -1, then i^4 = i² * i² = (-1) * (-1) = 1.

Looks quite coherent (to me).

Richard Hachel

2025-02-27 16:12:06 UTC

Your new "system" with i² = -1 and i^4 = -1 is not even consistent.

Indeed, especially if i^4 = i² * i².
After all, if i² = -1, then i^4 = i² * i² = (-1) * (-1) = 1.
And 1 =/= -1.

It is the system i² * i² = i^4 = 1 that does not seem coherent to me.

Strange.
After all, if i² = -1, then i^4 = i² * i² = (-1) * (-1) = 1.
Looks quite coherent (to me).

It is consistent if we do operations with real numbers.
The problem with mathematicians is that they propose a system where
i²=-1, which is no longer consistent with traditional operations, and
which is a new imaginary system (which I understand, and which I admit as
consistent and useful in itself), but that once this is accepted, they
return to classical formulations such as a²*a²=a^4>0.
"The sow that was washed and perfumed went back to wallowing in the mire".

R.H.

Chris M. Thomasson

2025-02-27 23:51:36 UTC

Your new "system" with i² = -1 and i^4 = -1 is not even consistent.

Indeed, especially if i^4 = i² * i².
After all, if i² = -1, then i^4 = i² * i² = (-1) * (-1) = 1.
And 1 =/= -1.

It is the system i² * i² = i^4 = 1 that does not seem coherent to me.

Strange.
After all, if i² = -1, then i^4 = i² * i² = (-1) * (-1) = 1.
Looks quite coherent (to me).

It is consistent if we do operations with real numbers.

How would you implement my Multi Julia algorithm using your system? It
might be fun to plot it... See how they look, then compare and contrast
the results:

http://www.paulbourke.net/fractals/multijulia

[...]

efji

2025-02-26 20:10:38 UTC

Post by Richard Hachel
It is the system i²*i²=i^4=1 that does not seem coherent to me.

To be able to write something like "i^4", which seems to mean in your
thick brain "i*i*i*i", you implicitly assume that your "*" is an
associative operator, otherwise "i^4" means nothing at all.

Then, if "*" is associative,
i^4 = i*i*i*i = i*(i*i)*i = (i*i)*(i*i) = (i*i)^2.
If you assume i^2 = i*i = -1, then i^4=1.

End of your pathetic story.
Now, please, shut the fuck up and go back to your psychiatric hospital.

--
F.J.

Richard Hachel

2025-02-26 20:22:01 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.

Just as we don't do subtractions with the rules of division, we don't
multiply imaginary numbers like crazy.
We have to respect the rules.
Of course, if we don't respect the natural rules, we're going to write
crap like i=-1, i²=-1, i^4=1...
And why not i^12=-(17/3)?
That's not very serious.

R.H.

Python

2025-02-26 20:26:09 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.
Just as we don't do subtractions with the rules of division, we don't multiply
imaginary numbers like crazy.

It is not "as crazy". It was explained in the part you've snipped.

You are a dishonest person, Richard. One of the worse I've ever met.

Post by Richard Hachel
We have to respect the rules.
Of course, if we don't respect the natural rules, we're going to write crap like
i=-1, i²=-1, i^4=1...

"the rules", "the natural rules"? ? Just because you said so? You are a
arrogant idiot, Richard, with NO integrity.

Post by Richard Hachel
And why not i^12=-(17/3)?
That's not very serious.

What is not serious is your idiotic contradictory claims. What is worse
are your lies.

Moebius

2025-02-27 06:56:13 UTC

Post by Richard Hachel
We have to respect the rules.
Of course, if we don't respect the [agreed upon] rules, we're going to write
crap like i = -1

He has a point there! :-P

After all, in this case we would get i*i = (-1) * (-1) = 1, while
playing by the rules [agreed upon in mathematics] we get i*i = -1.

Moebius

2025-02-27 07:51:23 UTC

Post by Richard Hachel
We have to respect the rules.
Of course, if we don't respect the [agreed upon] rules, we're going
to write crap like i = -1

He has a point there! :-P
After all, in this case we would get i*i = (-1) * (-1) = 1, while
playing by the rules [agreed upon in mathematics] we get i*i = -1 (as desired).

joes

2025-02-27 07:47:20 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.

-1*-1 is not 1?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

Richard Hachel

2025-02-27 15:42:29 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.

-1*-1 is not 1?

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.

R.H.

Python

2025-02-27 17:05:54 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.

-1*-1 is not 1?

Claro que si.
Pero, i*i=i²=-1 ; (i²)²=-1

This is obviously contradictory. (-1)*(-1) IS (i^2)*(i^2) if i^2 = -1. End
of Story.

Post by Richard Hachel
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.

In complex numbers it is FALSE that i = -1 and also FALSE that i(-1/2) =
-1
(i^(-1/2) = 1/sqrt(i) = 1/sqrt(2) - i/sqrt(2)) (considering principal
value
only, there is another which is the opposite)

Stop making up stuff, you are not only an idiot but also a liar.

Jim Burns

2025-02-27 19:35:58 UTC

Post by Richard Hachel
It seems that the imaginary unit i is
a special unit such that i^x=-1 whatever x.

Multiplication by 𝑖 is a ¹/₄.turn of the complex plane.

Multiplication twice by 𝑖 is a ¹/₂.turn, which is
multiplication by -1

The four solutions to x⁴ = -1 are
a ¹/₈.turn, a ³/₈.turn, a ⁵/₈.turn, and a ⁷/₈.turn
Repeated four times, they are
a ¹/₂.turn, a 1¹/₂.turn, a 2¹/₂.turn, and a 3¹/₈.turn,
which are multiplication by -1, -1, -1, and -1.

Multiplication by 3 is left as an exercise for the reader.

There is a very nice formula expressing all this.
𝑒ⁱᶿ = sin(θ) + 𝑖⋅cos(θ)

x⁴ = -81
x ∈ { 3𝑒ⁱᐧᵖⁱᐟ⁴, 3𝑒³ⁱᐧᵖⁱᐟ⁴, 3𝑒⁵ⁱᐧᵖⁱᐟ⁴, 3𝑒⁷ⁱᐧᵖⁱᐟ⁴ }

Jim Burns

2025-02-27 20:53:19 UTC

Post by Richard Hachel
It seems that the imaginary unit i is
a special unit such that i^x=-1 whatever x.

Multiplication by 𝑖 is a ¹/₄.turn of the complex plane.
Multiplication twice by 𝑖 is a ¹/₂.turn, which is
multiplication by -1
The four solutions to x⁴ = -1 are
a ¹/₈.turn, a ³/₈.turn, a ⁵/₈.turn, and a ⁷/₈.turn
Repeated four times, they are
a ¹/₂.turn, a 1¹/₂.turn, a 2¹/₂.turn, and a 3¹/₈.turn,
which are multiplication by -1, -1, -1, and -1.
Multiplication by 3 is left as an exercise for the reader.
There is a very nice formula expressing all this.
𝑒ⁱᶿ = sin(θ) + 𝑖⋅cos(θ)

D'oh!
𝑒ⁱᶿ = cos(θ) + 𝑖⋅sin(θ)

Post by Jim Burns
x⁴ = -81
x ∈ { 3𝑒ⁱᐧᵖⁱᐟ⁴, 3𝑒³ⁱᐧᵖⁱᐟ⁴, 3𝑒⁵ⁱᐧᵖⁱᐟ⁴, 3𝑒⁷ⁱᐧᵖⁱᐟ⁴ }

Richard Hachel

2025-02-27 21:42:10 UTC

Post by Jim Burns
There is a very nice formula expressing all this.
𝑒ⁱᶿ = sin(θ) + 𝑖⋅cos(θ)

D'oh!
𝑒ⁱᶿ = cos(θ) + 𝑖⋅sin(θ)

Yes, it's more better. :))

R.H.

Chris M. Thomasson

2025-02-27 23:55:24 UTC

Post by Jim Burns
There is a very nice formula expressing all this.
𝑒ⁱᶿ = sin(θ) + 𝑖⋅cos(θ)

D'oh!
𝑒ⁱᶿ = cos(θ) + 𝑖⋅sin(θ)

Yes, it's more better. :))

To gain n-ary roots of a target complex number, keep in mind that the
power is a signed integer here:
________________________
ct_complex
root_calc(
ct_complex const& z,
int p,
int n
) {
ct_real radius = std::pow(std::abs(z), 1.f / p);
ct_real angle_base = std::arg(z) / p;
ct_real angle_step = CT_PI2 / p;
ct_real angle = angle = angle_step * n;

ct_complex c = {
std::cos(angle_base + angle) * radius,
std::sin(angle_base + angle) * radius
};

return c;
}
________________________

Chris M. Thomasson

2025-02-27 23:53:25 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.

-1*-1 is not 1?

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?

Richard Hachel

2025-02-28 00:50:30 UTC

Post by Richard Hachel
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.

sobriquet

2025-02-28 01:40:09 UTC

Post by Richard Hachel
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.

You claim that -1 * -1 = -1
since you claim that i^4 = -1 = i^2 * i^2 = -1 * -1

Richard Hachel

2025-02-28 01:49:16 UTC

Post by sobriquet
Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2.

Ce n'est pas ce que j'ai dit.

Post by sobriquet
You claim that -1 * -1 = -1

Ce n'est pas non plus ce que j'ai dit.

Post by sobriquet
since you claim that i^4 = -1 = i^2 * i^2

Oui, ça, je l'ai dit.

= -1 * -1

Non.

R.H.

sobriquet

2025-02-28 02:24:08 UTC

Post by sobriquet
Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2.

Ce n'est pas ce que j'ai dit.

Post by sobriquet
You claim that -1 * -1 = -1

Ce n'est pas non plus ce que j'ai dit.

Post by sobriquet
since you claim that i^4 = -1 = i^2 * i^2

Oui, ça, je l'ai dit.
= -1 * -1
Non.
R.H.

Ok, so which of the following statements are false according to you?

i^4 = -1

i^2 = -1

i^2 = i * i

i^4 = i * i * i * i

i^4 = i^2 * i^2

Richard Hachel

2025-02-28 13:05:21 UTC

Post by sobriquet
Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2.

Ce n'est pas ce que j'ai dit.

Post by sobriquet
You claim that -1 * -1 = -1

Ce n'est pas non plus ce que j'ai dit.

Post by sobriquet
since you claim that i^4 = -1 = i^2 * i^2

Oui, ça, je l'ai dit.
= -1 * -1
Non.
R.H.

Ok, so which of the following statements are false according to you?
i^4 = -1

C'est vrai.

Post by sobriquet
i^2 = -1

C'est vrai.

Post by sobriquet
i^2 = i * i

C'est vrai.

Post by sobriquet
i^4 = i * i * i * i

C'est vrai.

Post by sobriquet
i^4 = i^2 * i^2

C'est vrai.

R.H.

sobriquet

2025-02-28 13:16:58 UTC

Post by sobriquet
Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2.

Ce n'est pas ce que j'ai dit.

Post by sobriquet
You claim that -1 * -1 = -1

Ce n'est pas non plus ce que j'ai dit.

Post by sobriquet
since you claim that i^4 = -1 = i^2 * i^2

Oui, ça, je l'ai dit.
= -1 * -1
Non.
R.H.

Ok, so which of the following statements are false according to you?
i^4 = -1

C'est vrai.

Post by sobriquet
i^2 = -1

C'est vrai.

Post by sobriquet
i^2 = i * i

C'est vrai.

Post by sobriquet
i^4 = i * i * i * i

C'est vrai.

Post by sobriquet
i^4 = i^2 * i^2

C'est vrai.
R.H.

So basically you're saying

-1 * -1 = -1

Because that is the way the equality relation works in math.

Richard Hachel

2025-02-28 13:26:50 UTC

Post by sobriquet
Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2.

Ce n'est pas ce que j'ai dit.

Post by sobriquet
You claim that -1 * -1 = -1

Ce n'est pas non plus ce que j'ai dit.

Post by sobriquet
since you claim that i^4 = -1 = i^2 * i^2

Oui, ça, je l'ai dit.
= -1 * -1
Non.
R.H.

Ok, so which of the following statements are false according to you?
i^4 = -1

C'est vrai.

Post by sobriquet
i^2 = -1

C'est vrai.

Post by sobriquet
i^2 = i * i

C'est vrai.

Post by sobriquet
i^4 = i * i * i * i

C'est vrai.

Post by sobriquet
i^4 = i^2 * i^2

C'est vrai.
R.H.

So basically you're saying
-1 * -1 = -1

No.

Post by sobriquet
Because that is the way the equality relation works in math.

No.

Not in complex mathematics.

If you want to use complex numbers, you have to use the laws of complex
numbers.
So you have to use i^x=-1 whatever x is.
Otherwise, it's like eating soup with a fork.

R.H.

efji

2025-02-28 13:29:46 UTC

Post by Richard Hachel
If you want to use complex numbers

Please, stop using the term "complex" for your shitty delirium. Thanks.

--
F.J.

Richard Hachel

2025-02-28 13:55:21 UTC

Post by Richard Hachel
If you want to use complex numbers

Please, stop using the term "complex" for your shitty delirium. Thanks.

Richard Hachel

2025-02-28 13:57:29 UTC

Post by Richard Hachel
If you want to use complex numbers

Please, stop using the term "complex" for your shitty delirium. Thanks.

I refuse.
I call a complex number an imaginary number that involves the imaginary
unit i.
And I don't see why I should be forbidden from voting for the imaginary
candidate I have chosen.

R.H.

Jim Burns

2025-02-28 14:17:49 UTC

Post by Richard Hachel
If you want to use complex numbers

Please, stop using the term "complex"
for your shitty delirium. Thanks.

I refuse.
I call a complex number
an imaginary number that involves
the imaginary unit i.
And I don't see why
I should be forbidden from voting for
the imaginary candidate I have chosen.

https://en.wikipedia.org/wiki/Cooperative_principle
⎛
⎜ These describe specific rational principles observed
⎜ by people who follow the cooperative principle
⎜ in pursuit of effective communication.
⎜ Applying the Gricean maxims is therefore
⎜ a way to explain the link between utterances and
⎝ what is understood from them.

Moebius

2025-02-28 14:26:09 UTC

Post by Richard Hachel
If you want to use complex numbers

Please, stop using the term "complex" for your shitty delirium. Thanks.

I suggest /delusional/ instead.

Richard Hachel

2025-02-28 14:32:12 UTC

Post by Richard Hachel
If you want to use complex numbers

Please, stop using the term "complex" for your shitty delirium. Thanks.

I suggest /delusional/ instead.

Shut up.

When you are up to it you will be able to speak.

And that is far from being your case.

You're a buffoon like the others.

A small shrill bell carried by the wind.

R.H

Moebius

2025-02-28 14:24:40 UTC

Post by sobriquet
So basically you're saying
-1 * -1 = -1

From this we get (with division by -1 on both sides):

-1 = 1.

And from this (with addition of 1 on both sides):

0 = 2 .

Division by 2 gives:

0 = 1 .

That's a fine result!

Moreover it implies

1 * 1 = 0 * 0 = 0 = 2 .

Just as you said:

"Your approach to math closely resembles Terrence Howard who claims

1 * 1 = 2."

Chris M. Thomasson

2025-02-28 18:29:24 UTC

Post by sobriquet
So basically you're saying
-1 * -1 = -1

-1 = 1.
0 = 2 .
0 = 1 .
That's a fine result!
Moreover it implies
1 * 1 = 0 * 0 = 0 = 2 .
"Your approach to math closely resembles Terrence Howard who claims
1 * 1 = 2."

Ditto.

Chris M. Thomasson

2025-02-28 01:50:17 UTC

Post by efji
If you assume i^2 = i*i = -1, then i^4=1.

Absolutely not.

-1*-1 is not 1?

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?

A fun aspect. I think there is a way to describe 2-ary complex numbers
on the real line using some special blocks, so to speak...

Jim Burns

2025-02-26 23:08:01 UTC

Post by Richard Hachel
But the error will quickly be found
if we use the unit i
as it must be used in this case.

We want a 2.dimensional field which extends
the 1.dimensional field of the real numbers.
That is to say, we want 2.dimensional
addition '+' and multiplication '⋅'
which satisfy the same laws satisfied by
our 1.dimensional '+' and '⋅':
⎛ associativity and commutativity for both,
⎜ identities 𝟎 𝟏, inverses -𝐱 𝐱⁻¹ except 𝟎⁻¹,
⎝ distributivity of '.' over '+'

We have what we want if
we have a vector 𝐯 not on the real axis such that,
for this 2.dimensional multiplication '⋅'
𝟏⋅𝟏 = 𝟏
𝟏⋅𝐯 = 𝐯
𝐯⋅𝟏 = 𝐯
𝐯⋅𝐯 = -α𝟏-2β𝐯
such that α > β²

Pick 𝐯 ∈ ℝ×(ℝ\{0}), β, α > β²
Define
𝐯⋅𝐯 = -α𝟏-2β𝐯
(a𝟏+b𝐯)⋅(c𝟏+d𝐯) = ac𝟏+(ad+bd)𝐯+bd(𝐯⋅𝐯)

We have what we want,
a 2.dimensional field extending ℝ

However,
suppose 𝐯 ≠ ⟨0,1⟩ and 𝐯⋅𝐯 ≠ -𝟏
Then 𝐯 ≠ 𝐢

But 𝐢 still exists,
and it's determined by choices 𝐯, β, α
𝐢 = ±(𝐯+β𝟏)/(α-β²)¹ᐟ² (either sign works)
𝐯 = ±(α-β²)¹ᐟ²𝐢-β𝟏

Each a𝟏+b𝐯 has a corresponding a′𝟏+b′𝐢
and vice versa.
𝟏⋅𝟏 = 𝟏
𝟏⋅𝐢 = 𝐢
𝐢⋅𝟏 = 𝐢
𝐢⋅𝐢 = -𝟏
(a𝟏+b𝐢)⋅(c𝟏+d𝐢) = (ac-bd)𝟏+(ad+bd)𝐢
in the usual way.

Post by Richard Hachel
The other four roots being incorrect
(in the proposed system).

Your proposed system does not have
2.dimensional '+' and '⋅' satisfying:
⎛ associativity and commutativity for both,
⎜ identities 𝟎 𝟏, inverses -𝐱 𝐱⁻¹ except 𝟎⁻¹,
⎝ distributivity of '.' over '+'

Chris M. Thomasson

2025-02-25 21:37:54 UTC

Post by Richard Hachel
x^4=-81
What is x?

Think of how many complex roots that when raised to the 4'th power equal
-81.

Chris M. Thomasson

2025-02-25 21:49:47 UTC

Post by Richard Hachel
x^4=-81
What is x?

Try to forgive the floating point precision aspects, but, the roots are
the r's, raising them to the 4'th power gives the p's:
_____________________
r0 = (2.12132,2.12132)
r1 = (-2.12132,2.12132)
r2 = (-2.12132,-2.12132)
r3 = (2.12132,-2.12132)

p0 = (-81,-7.08124e-06)
p1 = (-81,-1.93183e-06)
p2 = (-81,-7.53158e-05)
p3 = (-81,4.57051e-05)
_____________________

To gain a root, here is my code:
_____________________
ct_complex
root_calc(
ct_complex const& z,
int p,
int n
) {
float radius = std::pow(std::abs(z), 1.0 / p);
float angle_base = std::arg(z) / p;
float angle_step = (CT_PI * 2.0) / p;
float angle = angle = angle_step * n;

ct_complex c = {
std::cos(angle_base + angle) * radius,
std::sin(angle_base + angle) * radius
};

return c;
}
_____________________

Also, this is not using floating point for roots, just signed integers.

Chris M. Thomasson

2025-02-25 22:00:01 UTC

Post by Richard Hachel
x^4=-81
What is x?

Try to forgive the floating point precision aspects, but, the roots are
_____________________
r0 = (2.12132,2.12132)
r1 = (-2.12132,2.12132)
r2 = (-2.12132,-2.12132)
r3 = (2.12132,-2.12132)
p0 = (-81,-7.08124e-06)
p1 = (-81,-1.93183e-06)
p2 = (-81,-7.53158e-05)
p3 = (-81,4.57051e-05)
_____________________
_____________________
ct_complex
root_calc(
    ct_complex const& z,
    int p,
    int n
) {
    float radius = std::pow(std::abs(z), 1.0 / p);
    float angle_base = std::arg(z) / p;
    float angle_step = (CT_PI * 2.0) / p;
    float angle = angle = angle_step * n;

^^^^^^^^^^^^^^^^^^^^^^

It still compiles and works wrt C++, but this looks odd to me. I typed
this in the newsreader just to see if I could remember it.

That line should be:

float angle = angle_step * n;

Post by Chris M. Thomasson
    ct_complex c = {
        std::cos(angle_base + angle) * radius,
        std::sin(angle_base + angle) * radius
    };
    return c;
}
_____________________
Also, this is not using floating point for roots, just signed integers.

Richard Hachel

2025-02-25 22:09:31 UTC

Post by Richard Hachel
x^4=-81
What is x?

Try to forgive the floating point precision aspects, but, the roots are
_____________________
r0 = (2.12132,2.12132)
r1 = (-2.12132,2.12132)
r2 = (-2.12132,-2.12132)
r3 = (2.12132,-2.12132)
p0 = (-81,-7.08124e-06)
p1 = (-81,-1.93183e-06)
p2 = (-81,-7.53158e-05)
p3 = (-81,4.57051e-05)
_____________________
_____________________
ct_complex
root_calc(
ct_complex const& z,
int p,
int n
) {
float radius = std::pow(std::abs(z), 1.0 / p);
float angle_base = std::arg(z) / p;
float angle_step = (CT_PI * 2.0) / p;
float angle = angle = angle_step * n;
ct_complex c = {
std::cos(angle_base + angle) * radius,
std::sin(angle_base + angle) * radius
};
return c;
}
_____________________
Also, this is not using floating point for roots, just signed integers.

This is quite complicated, where I propose to use the nature of the
imaginary number i in a somewhat particular way, and according to the new
idea that i is not only defined by i²=-1 or i=sqrt(-1), but rather with
the generalized idea that for all x, i^x=-1.
A bit like if this imaginary was the antithesis of 1 where for all x, then
1^x=1.

With this technique, we immediately have x=3i.

R.H.

Chris M. Thomasson

2025-02-25 23:29:21 UTC

Post by Richard Hachel
x^4=-81
What is x?

Try to forgive the floating point precision aspects, but, the roots
_____________________
r0 = (2.12132,2.12132)
r1 = (-2.12132,2.12132)
r2 = (-2.12132,-2.12132)
r3 = (2.12132,-2.12132)
p0 = (-81,-7.08124e-06)
p1 = (-81,-1.93183e-06)
p2 = (-81,-7.53158e-05)
p3 = (-81,4.57051e-05)
_____________________
_____________________
ct_complex
root_calc(
     ct_complex const& z,
     int p,
     int n
) {
     float radius = std::pow(std::abs(z), 1.0 / p);
     float angle_base = std::arg(z) / p;
     float angle_step = (CT_PI * 2.0) / p;
     float angle = angle = angle_step * n;
     ct_complex c = {
         std::cos(angle_base + angle) * radius,
         std::sin(angle_base + angle) * radius
     };
     return c;
}
_____________________
Also, this is not using floating point for roots, just signed integers.

This is quite complicated, where I propose to use the nature of the
imaginary number i in a somewhat particular way, and according to the
new idea that i is not only defined by i²=-1 or i=sqrt(-1), but rather
with the generalized idea that for all x, i^x=-1.
A bit like if this imaginary was the antithesis of 1 where for all x,
then 1^x=1.
With this technique, we immediately have x=3i.

I don't know what you mean here?

Chris M. Thomasson

2025-02-25 23:31:08 UTC

Post by Richard Hachel
x^4=-81
What is x?

This is quite complicated,

Actually, it's quite straightforward wrt getting at the roots for a
target number. Even using signed integers for the power.

Post by Richard Hachel
where I propose to use the nature of the
imaginary number i in a somewhat particular way, and according to the
new idea that i is not only defined by i²=-1 or i=sqrt(-1), but rather
with the generalized idea that for all x, i^x=-1.
A bit like if this imaginary was the antithesis of 1 where for all x,
then 1^x=1.
With this technique, we immediately have x=3i.
R.H.

Chris M. Thomasson

2025-02-25 22:08:03 UTC

Post by Richard Hachel
x^4=-81
What is x?

what about x^7 = -81 ?

Using my own code:

r0 = (1.68791,0.812857)
r1 = (0.41688,1.82647)
r2 = (-1.16807,1.46472)
r3 = (-1.87344,-1.63782e-07)
r4 = (-1.16807,-1.46472)
r5 = (0.416881,-1.82647)
r6 = (1.68792,-0.812856)

p0 = (-81,-2.63931e-05)
p1 = (-81,-1.93183e-06)
p2 = (-81,2.25295e-05)
p3 = (-81,-4.69908e-05)
p4 = (-81,-0.000177025)
p5 = (-81,-7.53158e-05)
p6 = (-81,-0.000263285)

Richard Hachel

2025-02-25 22:20:57 UTC

Post by Richard Hachel
x^4=-81
What is x?

what about x^7 = -81 ?
r0 = (1.68791,0.812857)
r1 = (0.41688,1.82647)
r2 = (-1.16807,1.46472)
r3 = (-1.87344,-1.63782e-07)
r4 = (-1.16807,-1.46472)
r5 = (0.416881,-1.82647)
r6 = (1.68792,-0.812856)
p0 = (-81,-2.63931e-05)
p1 = (-81,-1.93183e-06)
p2 = (-81,2.25295e-05)
p3 = (-81,-4.69908e-05)
p4 = (-81,-0.000177025)
p5 = (-81,-7.53158e-05)
p6 = (-81,-0.000263285)

x=-1.873444

R.H.

Chris M. Thomasson

2025-02-25 22:53:49 UTC

Post by Richard Hachel
x^4=-81
What is x?

x=-1.873444

The n-ary roots for (-1.873444+0i)?

here is the output from my program wrt the 4'th power:

roots[0] = (0.827266,0.827266)
roots[1] = (-0.827266,0.827266)
roots[2] = (-0.827266,-0.827266)
roots[3] = (0.827266,-0.827266)

raised[0] = (-1.87344,-1.63782e-07)
raised[1] = (-1.87344,-4.46812e-08)
raised[2] = (-1.87344,-1.74197e-06)
raised[3] = (-1.87344,1.05711e-06)

Chris M. Thomasson

2025-02-25 23:43:36 UTC

Post by Richard Hachel
x^4=-81
What is x?

x=-1.873444

[...]

Extending the precision a bit:

roots[0] = (0.827266,0.827266)
roots[1] = (-0.827266,0.827266)
roots[2] = (-0.8272659,-0.8272662)
roots[3] = (0.8272659,-0.8272662)

raised[0] = (-1.873444,-1.637816e-07)
raised[1] = (-1.873444,-4.468118e-08)
raised[2] = (-1.873444,-1.741974e-06)
raised[3] = (-1.873444,1.05711e-06)

Chris M. Thomasson

2025-02-25 23:48:08 UTC

Post by Richard Hachel
x^4=-81
What is x?

[...]

Post by Richard Hachel
x=-1.873444

To the 13'th power with higher precision:

roots[0] = (1.01898,0.251156)
roots[1] = (0.7855438,0.6959311)
roots[2] = (0.3721492,0.9812768)
roots[3] = (-0.1265003,1.041824)
roots[4] = (-0.5961701,0.8637015)
roots[5] = (-0.9292645,0.4877156)
roots[6] = (-1.049476,5.945845e-16)
roots[7] = (-0.9292645,-0.4877156)
roots[8] = (-0.5961701,-0.8637015)
roots[9] = (-0.1265003,-1.041824)
roots[10] = (0.3721492,-0.9812768)
roots[11] = (0.7855438,-0.6959311)
roots[12] = (1.01898,-0.251156)

raised[0] = (-1.873444,2.294307e-16)
raised[1] = (-1.873444,4.016197e-15)
raised[2] = (-1.873444,4.475059e-15)
raised[3] = (-1.873444,1.606015e-15)
raised[4] = (-1.873444,2.064877e-15)
raised[5] = (-1.873444,9.179548e-15)
raised[6] = (-1.873444,9.63841e-15)
raised[7] = (-1.873444,4.132072e-15)
raised[8] = (-1.873444,4.590934e-15)
raised[9] = (-1.873444,1.170561e-14)
raised[10] = (-1.873444,2.214818e-14)
raised[11] = (-1.873444,1.262333e-14)
raised[12] = (-1.873444,2.306591e-14)

Richard Hachel

2025-02-26 00:11:36 UTC

Post by Richard Hachel
x^4=-81
What is x?

[...]

Post by Richard Hachel
x=-1.873444

roots[0] = (1.01898,0.251156)
roots[1] = (0.7855438,0.6959311)
roots[2] = (0.3721492,0.9812768)
roots[3] = (-0.1265003,1.041824)
roots[4] = (-0.5961701,0.8637015)
roots[5] = (-0.9292645,0.4877156)
roots[6] = (-1.049476,5.945845e-16)
roots[7] = (-0.9292645,-0.4877156)
roots[8] = (-0.5961701,-0.8637015)
roots[9] = (-0.1265003,-1.041824)
roots[10] = (0.3721492,-0.9812768)
roots[11] = (0.7855438,-0.6959311)
roots[12] = (1.01898,-0.251156)
raised[0] = (-1.873444,2.294307e-16)
raised[1] = (-1.873444,4.016197e-15)
raised[2] = (-1.873444,4.475059e-15)
raised[3] = (-1.873444,1.606015e-15)
raised[4] = (-1.873444,2.064877e-15)
raised[5] = (-1.873444,9.179548e-15)
raised[6] = (-1.873444,9.63841e-15)
raised[7] = (-1.873444,4.132072e-15)
raised[8] = (-1.873444,4.590934e-15)
raised[9] = (-1.873444,1.170561e-14)
raised[10] = (-1.873444,2.214818e-14)
raised[11] = (-1.873444,1.262333e-14)
raised[12] = (-1.873444,2.306591e-14)

I think that for the moment, we are making things terribly complicated.
If I ask you the cube root of 27?
Are you going to make a computer program?
Why make a computer program if I ask you the fourth root of -81?

The answer is simple and obvious. x=3i.

All these misunderstandings come from the fact that no clear and
universally usable definition of the imaginary number i has ever been
given.

Against all expectations, in analytical mathematics, i is an imaginary
unit such that, for all x, i^x=-1.

We see that saying that i²=-1 is completely legal.

Or that sqrt(i)=i^(1/2)=-1.

Certainly.

But we also see that (i²)² is not equal to 1, and that those who believe
it are corrupting themselves.

R.H.

Chris M. Thomasson

2025-02-26 00:34:58 UTC

Post by Richard Hachel
x^4=-81
What is x?

[...]

Post by Richard Hachel
x=-1.873444

The fourth root of -81+0i wrt power of 4 is *:

roots[0] = (2.12132,2.12132)
roots[1] = (-2.12132,2.12132)
roots[2] = (-2.12132,-2.121321)
*roots[3] = (2.12132,-2.121321)

I don't know what you x=3i even means right now. Any of these roots
raised to the 4'th power equals -81+0i.

Post by Richard Hachel
All these misunderstandings come from the fact that no clear and
universally usable definition of the imaginary number i has ever been
given.
Against all expectations, in analytical mathematics, i is an imaginary
unit such that, for all x, i^x=-1.
We see that saying that i²=-1 is completely legal.
Or that sqrt(i)=i^(1/2)=-1.
Certainly.
But we also see that (i²)² is not equal to 1, and that those who believe
it are corrupting themselves.
R.H.

Richard Hachel

2025-02-26 01:11:18 UTC

Post by Chris M. Thomasson
roots[0] = (1.01898,0.251156)
roots[1] = (0.7855438,0.6959311)
roots[2] = (0.3721492,0.9812768)
roots[3] = (-0.1265003,1.041824)
roots[4] = (-0.5961701,0.8637015)
roots[5] = (-0.9292645,0.4877156)
roots[6] = (-1.049476,5.945845e-16)
roots[7] = (-0.9292645,-0.4877156)
roots[8] = (-0.5961701,-0.8637015)
roots[9] = (-0.1265003,-1.041824)
roots[10] = (0.3721492,-0.9812768)
roots[11] = (0.7855438,-0.6959311)
roots[12] = (1.01898,-0.251156)
raised[0] = (-1.873444,2.294307e-16)
raised[1] = (-1.873444,4.016197e-15)
raised[2] = (-1.873444,4.475059e-15)
raised[3] = (-1.873444,1.606015e-15)
raised[4] = (-1.873444,2.064877e-15)
raised[5] = (-1.873444,9.179548e-15)
raised[6] = (-1.873444,9.63841e-15)
raised[7] = (-1.873444,4.132072e-15)
raised[8] = (-1.873444,4.590934e-15)
raised[9] = (-1.873444,1.170561e-14)
raised[10] = (-1.873444,2.214818e-14)
raised[11] = (-1.873444,1.262333e-14)
raised[12] = (-1.873444,2.306591e-14)

roots[0] = (2.12132,2.12132)
roots[1] = (-2.12132,2.12132)
roots[2] = (-2.12132,-2.121321)
*roots[3] = (2.12132,-2.121321)
I don't know what you x=3i even means right now. Any of these roots
raised to the 4'th power equals -81+0i.

We are not talking about the same thing, nor are we using the same
mathematics.

If I ask what are the complex roots of f(x)=x²+4x+5,
you will tell me that we must use [-b$sqrt(b²-4ac)]/2a using i.

And you will give me x'=-2+i and x"=-2-i.

Coordinates on x'Ox that I will immediately place in A(-3,0) and B(-1,0)
and which are the imaginary roots on y=0, the equation having no real
roots.

Well, I do the same to find the fourth root of -81.

x^4=-81.

For me, i ^x=-1 whatever 1.

x^4=-81 ---> x^4=-(i^4)(-81)=81(i^4)
x=3i

Conversely, x^4=(3i)^4 = 81(i^4) with i^4=-1 by definition of i for me.

But that has nothing to do with the Argand frame, which is something
completely different.

R.H.

Chris M. Thomasson

2025-02-26 01:59:11 UTC

We are not talking about the same thing, nor are we using the same
mathematics.
If I ask what are the complex roots of f(x)=x²+4x+5,
you will tell me that we must use [-b$sqrt(b²-4ac)]/2a using i.
And you will give me x'=-2+i and x"=-2-i.
Coordinates on x'Ox that I will immediately place in A(-3,0) and B(-1,0)
and which are the imaginary roots on y=0, the equation having no real
roots.
Well, I do the same to find the fourth root of -81.
x^4=-81.
For me, i ^x=-1 whatever 1.
x^4=-81 ---> x^4=-(i^4)(-81)=81(i^4)
x=3i

List out all of the four roots of -81+0i using your system such that
when they are raised to the 4'th power equal -81. I know how to find
them using good ol' complex numbers.

Post by Richard Hachel
Conversely, x^4=(3i)^4 = 81(i^4) with i^4=-1 by definition of i for me.
But that has nothing to do with the Argand frame, which is something
completely different.
R.H.

Chris M. Thomasson

2025-02-26 00:37:51 UTC

[...]

The 2 roots of 9+0i are, wrt power of 2, are:

roots[0] = (3,0)
roots[1] = (-3,-2.622683e-07)

Each one of these when raised to the 2'nd power is 9.

The two roots of 0+9i are:

roots[0] = (2.12132,2.12132)
roots[1] = (-2.12132,-2.121321)

Why do you seem to misunderstand complex numbers so much?

efji

2025-02-26 08:51:12 UTC

Post by Chris M. Thomasson
Why do you seem to misunderstand complex numbers so much?

Don't lose your time answering this disturbed and provocative person. He
mainly needs to be admitted to a psychiatric hospital. The sooner, the
better.

--
F.J.

Chris M. Thomasson

2025-02-27 23:56:41 UTC

Post by Chris M. Thomasson
Why do you seem to misunderstand complex numbers so much?

Don't lose your time answering this disturbed and provocative person. He
mainly needs to be admitted to a psychiatric hospital. The sooner, the
better.

Yeah. Damn... Well, humm... I am still wondering how he would implement
one of my algorithms using his "special" system:

http://www.paulbourke.net/fractals/multijulia/

Chris M. Thomasson

2025-02-26 00:39:35 UTC

Post by Richard Hachel
x^4=-81
What is x?

[...]

Post by Richard Hachel
x=-1.873444

I think that for the moment, we are making things terribly complicated.
If I ask you the cube root of 27?
Are you going to make a computer program?

Huh? I use complex roots all the time, and yes, I have programs for it.
How do you think I created my Multi Julia algorithm?

http://www.paulbourke.net/fractals/multijulia/

How would you implement my algo? I know how to do it for sure.

Post by Richard Hachel
Why make a computer program if I ask you the fourth root of -81?
The answer is simple and obvious. x=3i.
All these misunderstandings come from the fact that no clear and
universally usable definition of the imaginary number i has ever been
given.
Against all expectations, in analytical mathematics, i is an imaginary
unit such that, for all x, i^x=-1.
We see that saying that i²=-1 is completely legal.
Or that sqrt(i)=i^(1/2)=-1.
Certainly.
But we also see that (i²)² is not equal to 1, and that those who believe
it are corrupting themselves.
R.H.

Chris M. Thomasson

2025-02-26 00:46:43 UTC

Post by Richard Hachel
x^4=-81
What is x?

[...]

Post by Richard Hachel
x=-1.873444

I think that for the moment, we are making things terribly complicated.
If I ask you the cube root of 27?
Are you going to make a computer program?

Huh? I use complex roots all the time, and yes, I have programs for it.
How do you think I created my Multi Julia algorithm?
http://www.paulbourke.net/fractals/multijulia/
How would you implement my algo? I know how to do it for sure.

Btw, the cube root of 27+0i is:

roots[0] = (3,0)
roots[1] = (-1.5,2.598076)
roots[2] = (-1.5,-2.598076)

Barry Schwarz

2025-02-26 10:30:42 UTC

On Wed, 26 Feb 25 00:11:36 +0000, Richard Hachel <***@tiscali.fr>
wrote:

<snip>

Post by Richard Hachel
I think that for the moment, we are making things terribly complicated.

By we of course you mean your refusal to accept the standard meaning
of a complex number.

Post by Richard Hachel
If I ask you the cube root of 27?
Are you going to make a computer program?
Why make a computer program if I ask you the fourth root of -81?
The answer is simple and obvious. x=3i.

And incorrect. In standard mathematics, (3i)^4 does not equal -81

Post by Richard Hachel
All these misunderstandings come from the fact that no clear and
universally usable definition of the imaginary number i has ever been
given.

Strange that none of us seem to have a problem with the standard
definition that has been around for centuries.

Post by Richard Hachel
Against all expectations, in analytical mathematics, i is an imaginary
unit such that, for all x, i^x=-1.

Only in your imagination.

Post by Richard Hachel
We see that saying that i²=-1 is completely legal.
Or that sqrt(i)=i^(1/2)=-1.

Your sqrt function is also non-standard. In standard math, if
y = sqrt(x) then y^2 = (sqrt(x))^2 = x. In your case, saying
sqrt(i) = -1 yields i = (sqrt(i))^2 = (-1)^2 = 1 which, of course, you
refuse to accept.

Post by Richard Hachel
But we also see that (i²)² is not equal to 1, and that those who believe
it are corrupting themselves.

Considering that the standard definition works in physics,
electronics, and other sciences, they must be corrupt also. It's a
wonder that your computer works and you can access the internet
through all this corruption.

--
Remove del for email

Richard Hachel

2025-02-26 16:55:42 UTC

Post by Barry Schwarz

Post by Richard Hachel
The answer is simple and obvious. x=3i.

And incorrect. In standard mathematics, (3i)^4 does not equal -81

That's what I'm saying.

But for me, (3i)^4=-81 (if always i^x=-1 ).

I'm just trying to understand things clearly at the base.

Standard mathematics poses i²=-1; which is justified, and it helps to
solve quadratic equations without roots.

We then realize that the imaginary roots found for f(x) are the real roots
of another curve g(x) which, on a Cartesian plane is the mirror curve of
f(x) centered on the vertex of the other curve.

Thus, and conversely, the complex roots of one are always the real roots
of the other.

So far, it's very useful for this kind of small mathematical maneuver, and
very precise.

Now saying that i^x is -1 if x=2 is very good.

But I would like this very restrictive definition to be abandoned.

The goal is to define i^x whatever x, and the way it is done does not suit
me.

It would be interesting to propose something like i^x=-1 whatever x, and
in this sense make i invariant whatever the power we give it.

This would be a very important, general, and much more practical
definition.

R.H.

Moebius

2025-02-26 17:11:38 UTC

i is an imaginary unit such that, for all x, i^x=-1.

Fascinating! So i = i^1 = -1 in your "system"?

But we also see that (i²)² is not equal to 1,

Hmmmm... i² = -1. Hence (i²)² = (-1)² = 1, in standard math.

Sorry, Barry, but I've blocked this idiot.

Richard Hachel

2025-02-26 18:16:21 UTC

i is an imaginary unit such that, for all x, i^x=-1.

Fascinating! So i = i^1 = -1 in your "system"?

Absolutely.

Je ne me contente pas d'écrire i²=-1 mais pour tout x, i^x=-1.

Nous sommes en train d'utiliser une unité imaginaire et nous devons
l'utiliser correctement jusqu'à bout, et pas seulement si x=2.

But we also see that (i²)² is not equal to 1,

Hmmmm... i² = -1. Hence (i²)² = (-1)² = 1, in standard math.

Yes, in standard maths.

No, with me.

Post by Moebius
Sorry, Barry, but I've blocked this idiot.

Goodbye.

R.H.

Python

2025-02-26 18:31:11 UTC

i is an imaginary unit such that, for all x, i^x=-1.

Fascinating! So i = i^1 = -1 in your "system"?

Absolutely.
Je ne me contente pas d'écrire i²=-1 mais pour tout x, i^x=-1.
Nous sommes en train d'utiliser une unité imaginaire et nous devons l'utiliser
correctement jusqu'à bout, et pas seulement si x=2.

"imaginary" does not mean that you can accept contradiction this is not a
fairy tale.

Using such a word when it comes to complex numbers is an history relief
because they were first used, centuries ago, without proper definition.
Mathematician just noticed that "it works" but stayed doubtful before a
proper consistent definition was coined up.

The word "imaginary" could have been dropped out at that time. There is no
problem in preserving it as long as we know it has since then a proper
meaning (completely different to the usual sense). Moreover in math it is
common (contrary to physics) to use common words with different meanings
than the usual one (matrix, ring, field, etc.)

Mathematicians are NOT supposed to take care of stupid hypocrite kooks of
your kind. You don't matter Lengrand.

Barry Schwarz

2025-02-25 22:42:59 UTC

Post by Richard Hachel
x^4=-81
What is x?
R.H.

One of the values of x is 1.5*sqrt(2)+1.5*sqrt(2)*i. You can find the
others yourself if you are willing to use standard math.

--
Remove del for email

Chris M. Thomasson

2025-02-28 07:10:19 UTC