Re: Hello!

Le 20/01/2025 à 00:40, Jim Burns a écrit :

cOn 1/19/2025 2:41 AM, Richard Hachel wrote:

Le 19/01/2025 à 02:53, sobriquet a écrit :

For ⅈ=1 or ⅈ=-1,  ⅈ²=1 or ⅈ²=1

I answered this in the previous post.
I said that mathematicians confuse, when squaring i, the following two objects.
i²=|i²|
and i²= |i|²
I repeat, when we do not know z, which can have two values ​​simultaneously (as in my example of the Plougastel high school),
we must give i an imaginary value, and i is neither 1 nor -1, but both at the same time.
And its square becomes i²=(-1)(1)=-1
But as soon as we know it (we can photograph it when it is -1 or when it is 1), we must no longer consider it as a duplicate.
If it is -1, we must follow the idea through to the end:
(a+ib)(a'+ib')=aa'-(ab'+a'b)+bb'
Same for i=1
(a+ib)(a'+ib')=aa'+(ab'+a'b)+bb'
We then realize that for the real part, we always have,
A=aa'+bb' whatever the value given to i.
On the other hand, and here, let's reintroduce i to make only one of the two equations:
Z=aa'+i(ab'+a'b)+bb'
where i can resume its bipolarity without any problem, while aa'+bb' is a correct real part for the equation.
Do you understand these things?
R.H.