Re: x²+4x+5=0

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