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Subject: Re: Division of two complex numbers
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From: Python <jp@python.invalid>

Le 20/01/2025 à 22:45, "Chris M. Thomasson" a écrit :
> On 1/20/2025 1:35 PM, Python wrote:
>> Le 20/01/2025 à 22:28, "Chris M. Thomasson" a écrit :
>>> On 1/20/2025 1:09 PM, Python wrote:
>>>> 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.
>>>
>>> No. GLSL is not C at all, it has a similar style, but is different for 
>>> sure.
>> 
>> It is exactly the same syntax. *facepalm*.
>> Ok, let's say so, if you wish, so you can implement complex 
>> multiplication in a GLSL shader.
> 
> No. C and GLSL are completely different languages. Have you ever even 
> used GLSL? You can do fun things in GLSL that C cannot do at all.

This is ridiculous nitpicking. 

#define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)

may compile in C, also can your function ct_cmul above. 

I didn't wrote that GLSL was C, I wrote that the code you wrote was C.

Anyway, this is not the point. Either in C or GLSL the fact that you can 
implement complex multiplication (or in ANY language) is NOT THE POINT it 
is IRRELEVANT!

>> Again: SO WHAT? ? ? This is NOT THE POINT of the discussion.
> 
> I thought it might help the OP.

In what manner? ?  Nobody, not even the OP pretended that it cannot be 
implemented.

Seriously Chris, what's wrong with you?

>>>> Again what's *your* point? Your posts makes absolutely no sense in 
>>>> the context of this thread!
>>>
>>> Just a way to multiply two 2-ary vectors as if they were complex 
>>> numbers. Now, here is a little C99 program I just typed in the 
>>> newsreader. It should compile.
>>> _____________________________
>>> [snip irrelevant triviality]
>> 
>> So what? ? ?
>>> I thought it might help out the OP.
>> 
>> In which way? ? ? Hachel didn't write that it cannot be done (he's not 
>> that silly), he claimed (wrongly) that it is the wrong way to define 
>> multiplication between complex numbers.
>> 
>>>>>>
>>>>>> 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*.