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Subject: Re: Division of two complex numbers
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Date: Mon, 20 Jan 25 21:09:47 +0000
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Le 20/01/2025 à 22:06, "Chris M. Thomasson" a écrit :
> On 1/20/2025 1:04 PM, Python wrote:
>> Le 20/01/2025 à 21:59, "Chris M. Thomasson" a écrit :
>>> On 1/20/2025 12:51 PM, Python wrote:
>>>> Le 20/01/2025 à 21:44, "Chris M. Thomasson" a écrit :
>>>>> On 1/20/2025 12:20 PM, Python wrote:
>>>>>> 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 :
>>>>>>
>>>>>> (0, 1) ^ 2 = 0
>>>>>>
>>>>>>
>>>>>
>>>>> vec2 ct_cmul(in vec2 p0, in vec2 p1)
>>>>> {
>>>>>      return vec2(p0.x * p1.x - p0.y * p1.y, p0.x * p1.y + p0.y * p1.x);
>>>>> }
>>>>
>>>> So what? This is not an application of the binomial formula...
>>>>
>>>> What's you point?
>>>>
>>>>
>>>
>>> It's a way I multiply two vectors together as if they are complex 
>>> numbers.
>>>
>>> Another one:
>>>
>>> #define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)
>>>
>>> I can pass in normal vectors to this in GLSL. vec2's
>> 
>> Good! You know how to write a C program. :-) (pun intended)
> 
> Fwiw, that is not is C, it's from one of my GLSL shaders. ;^)

It is also C. 

Again what's *your* point? Your posts makes absolutely no sense in the 
context of this thread!

>> 
>> This is quite off-topic to point out that multiplication of complex 
>> numbers in C/C++ can be done.
>> 
>> The discussion is not about that it can be done, even crank Hachel would 
>> admit this. It is *why* it makes sense to define multiplication *that way*.