Re: Equation complexe

Le 26/02/2025 à 18:57, Jim Burns a écrit :

On 2/25/2025 3:07 PM, Richard Hachel wrote:

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

 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.