Am 22.01.2025 um 23:16 schrieb Chris M. Thomasson:

On 1/22/2025 5:48 AM, Richard Hachel wrote:

x² + 4x + 5 = 0

This equation has no root, and <bla>

It has the two "roots" (solutions):

x = -2 - i
x = -2 + i

Hint:

x² + 4x + 5 = (x + 2)² + 1.

Hence x² + 4x + 5 = 0 is equivalent with (x + 2)² + 1 = 0 or (x + 2)² =
-1. So x + 2 has to be a (one or more) complex number(s) z such that z²
= -1. We know such numbers, they are i and -i (and there aren't more).
Hence we have x + 2 = i or x + 2 = -i resp. And hence x = -2 + i or x =
-2 - i resp.

The x^2 component has two roots.

Doesn't make much sense. :-P

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Am 22.01.2025 um 23:42 schrieb Moebius:

Am 22.01.2025 um 23:16 schrieb Chris M. Thomasson:

On 1/22/2025 5:48 AM, Richard Hachel wrote:

x² + 4x + 5 = 0

This equation has no root, and <bla>

Hint: "Every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots."

See: https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

It has the two "roots" (solutions):

x = -2 - i
x = -2 + i

Hint:

x² + 4x + 5 = (x + 2)² + 1.

Hence x² + 4x + 5 = 0 is equivalent with (x + 2)² + 1 = 0 or (x + 2)² = -1. So x + 2 has to be a (one or more) complex number(s) z such that z²
= -1. We know such numbers, they are i and -i (and there aren't more).
Hence we have x + 2 = i or x + 2 = -i resp. And hence x = -2 + i or x =
-2 - i resp.

The x^2 component has two roots.

Doesn't make much sense. :-P

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Am 22.01.2025 um 23:57 schrieb Chris M. Thomasson:

The x^2 component has two roots.

Doesn't make much sense. :-P

Shit happens. When I see x^2 and think of complex numbers, I think of
two roots. This can imply that there are two solutions.

As I said, doesn't make much sense.

Hint: An equation like x^2 = c (with c e IR or c e C \ {R}) has "roots"
(i.e. solutions).

Some examples:

x^2 = 0

x = 0.

x^2 = 1

x = 1 or x = -1.

x^2 = -1

x = i or x = -1.

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Am 23.01.2025 um 00:22 schrieb Chris M. Thomasson:

On 1/22/2025 3:20 PM, Moebius wrote:

Am 23.01.2025 um 00:11 schrieb Chris M. Thomasson:

On 1/22/2025 3:02 PM, Chris M. Thomasson wrote:

x^3+1+x^2 = 0

Wrt x^3,

<sigh>

The equations x^3 + x^2 + 1 = 0 has 3 "roots" (solutions).

I thought I wrote that.

No, you didn't.

Now, shut up, you idiot!

EOD

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Am 23.01.2025 um 00:22 schrieb Chris M. Thomasson:

A real is any complex number with a zero imaginary part?

Hint: x + i0 = x + 0 = x for all x e IR.

Yeah there's a subtle "problem":

(x, 0) = x

can't "actually" be right. But in math we "identify" (x, 0) with x. So,
yes, "a real is any complex number with a zero imaginary part" (sort of).

When we use the usual notation z = x + iy (where x,y e IR) for z e C,
the "propblem" vanishes.

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Am 23.01.2025 um 00:22 schrieb Chris M. Thomasson:

A real is any complex number with a zero imaginary part?

Hint: x + i0 = x + 0 = x for all x e IR.

Yeah, there's a subtle "problem":

(x, 0) = x

can't "actually" be right. But in math we "identify" (x, 0) with x. So,
yes, "a real is any complex number with a zero imaginary part" (sort of).

When we use the usual notation z = x + iy (where x,y e IR) for z e C,
the "propblem" vanishes.

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Am 23.01.2025 um 00:30 schrieb Moebius:

Am 23.01.2025 um 00:30 schrieb Chris M. Thomasson: <bla>

*plonk*

Bye.

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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

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.

Wolfram Alpha tells us there are two roots:

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

Here you can see the roots:

https://www.desmos.com/3d/mpwj5h2ab8

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Am 23.01.2025 um 00:11 schrieb Chris M. Thomasson:

On 1/22/2025 3:02 PM, Chris M. Thomasson wrote:

x^3+1+x^2 = 0

Wrt x^3,

<sigh>

The equations x^3 + x^2 + 1 = 0 has 3 "roots" (solutions).

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Am 23.01.2025 um 00:02 schrieb Chris M. Thomasson:

For instance, this should have three roots?

x^3 + x^2 + 1 = 0

Right.

One real and 2 complex solutions.

What am I missing here?

Nothing.

See: https://en.wikipedia.org/wiki/Cubic_equation

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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

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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

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.

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

.
.
.

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On 23/01/2025 11:01, Richard Hachel wrote:

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.

Ah, non - je l'ai mal interprete, j'ai lu comme x²+5x+5=0 :(


Peter F

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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."

:-)

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On 1/23/2025 3:34 AM, Richard Hachel wrote:

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

x²+4x+5=0

What is this imaginary mirror curve?
It is the curve with equation y=-x²-4x-3

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

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.

x²-2 = 0 has no solutions in the rational line.
x²-2 = 0 has two solutions in the complete (real) line.

x²+4x+5=0 has no solutions in the (real) line.
x²+4x+5=0 has two solutions in the (complex) plane.

x²+4x+5 = (x-(-2-𝑖))(x-(-2+𝑖)) for 𝑖²=-1

-x²-4x-3 = -1⋅(x-(-1))(x-(-3))

----
Why is 𝑖²=-1 ?

Define a plane.multiplication operator '∘': ℝ²×ℝ² → ℝ²
...such that '∘' is bilinear.
...such that left.identity and right.identity are
⎡1⎤
⎣0⎦
... such that each non.zero plane.point has a reciprocal.

Then, necessarily,
there is some λ > 0 and some μ such that
⎡a⎤∘⎡c⎤  :=
⎣b⎦ ⎣d⎦

⎛⎡⎡1 0⎤ ⎡0 -μ²-λ⎤⎤ ⎡a⎤⎞ ⎡c⎤  = ⎝⎣⎣0 1⎦,⎣1   2μ ⎦⎦ ⎣b⎦⎠ ⎣d⎦

⎡ac-(μ²+λ)bd⎤
⎣bc+ad+2μbd ⎦

Solve
⎡x⎤∘⎡x⎤  =  ⎡-1⎤ ⎣y⎦ ⎣y⎦     ⎣ 0⎦

x²-(μ²+λ)y² = -1
2xy+2μy² = 0

⎡x⎤  = ±⎡-μλ⁻¹ᐟ²⎤ ⎣y⎦     ⎣ λ⁻¹ᐟ² ⎦

Define
⎡-μλ⁻¹ᐟ²⎤   =:  𝑖
⎣ λ⁻¹ᐟ² ⎦

For that operation '∘': ℝ²×ℝ² → ℝ²
(a+b𝑖)∘(c+d𝑖) = (ac-bd)+(ad+bc)𝑖
𝑖∘𝑖 = -1
and
⟨ℝ²,+,∘⟩ is a field:
'+': ℝ²×ℝ² → ℝ² and '∘'
are associative and commutative,
have identities and inverses, except no 0⁻¹,
and '∘' distributes over '+'.

⎛ Side note:
⎜ Let (a+b𝑖)⃰ = a-b𝑖
⎜ Then
⎝ (w⃰∘z⃰)⃰ = w∘z

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