Re: Division of two complex numbers

Am 20.01.2025 23:08:44 Python schrieb:

Le 20/01/2025 à 23:00, Tom Bola a écrit :

Am 20.01.2025 21:20:51 Python schrieb:

Le 20/01/2025 à 21:09, Tom Bola a écrit :

Am 20.01.2025 20:33:12 Moebius schrieb:

Am 20.01.2025 um 19:27 schrieb Python:

Le 20/01/2025 à 19:23, Richard Hachel  a écrit :

Le 20/01/2025 à 19:10, Python a écrit :

Le 20/01/2025 à 18:58, Richard Hachel  a écrit :

Mathematicians give:
>
z1/z2=[(aa'+bb')/(a'²+b'²)]+i[(ba'-ab')/(a'²+b'²)]
>
It was necessary to write:
z1/z2=[(aa'-bb')/(a'²-b'²)]+i[(ba'-ab')/(a'²-b'²)]

>

I've explained how i is defined in a positive way in modern algebra.
i^2 = -1 is not a definition. It is a *property* that can be deduced
from a definition of i.

>
 That is what I saw.
>
 Is not a definition.
 It doesn't explain why.
>
We have the same thing with Einstein and relativity.
>
[snip unrelated nonsense about your idiotic views on Relativity]

 

It is clear that i²=-1, but we don't say WHY. It is clear however that
if i is both 1 and -1 (which gives two possible solutions) we can
consider its square as the product of itself by its opposite, and vice
versa.

 
I've posted a definition of i (which is NOT i^2 = -1) numerous times. A
"positive" definition as you asked for.

 
I've already told this idiot:
 
Complex numbers can be defined as (ordered) pairs of real numbers.
 
Then we may define (in this context):
 
          i := (0, 1) .
 
 From this we get: i^2 = -1.

 
For R.H.
  By the binominal formulas we have: (a, b)^2 = a^2 + 2ab + b^2

 
Huh? This is not the binomial formula which is (a + b)^2 = a^2 + 2ab + b^2
 
(a, b)^2 does not mean anything without any additional definition/context.
 

  So we get: (0, 1)^2 ) 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1

 
you meant  (0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1 ?
 
This does not make sense without additional context.
 
In R(epsilon) = R[X]/X^2 (dual numbers a + b*epsilon where epsilon is such
as
epsilon =/= 0 and epsilon^2 0) we do have :

 
So your epsilon or i squared is 0 --- but the complex i sqared is -1.

 
Sure. I didn't pretend that R(epsilon) was C. My point is that
multiplication in this ring makes perfect sense even if it is not
isomorphic to C.
 

(0, 1) ^ 2 = 0

 
But with your vector (a,b), b^2=-1 that vector square really does calculate
with the first binominal formula and you don't get " > (0, 1) ^ 2 = 0"...

 
The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about (0,
1)^2 or (a, b)^2.

Why not! ( Ok, you could also write (a + bi)^2 = -1,i^2 := -1 as well... )
 

When you wrote:
 
(0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1
----------------------^^^
 
where does that -1 comes from?

It does come from -2 + 1 = -1 as the solution of the line above...