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Subject: Re: Division of two complex numbers
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Date: Mon, 20 Jan 25 20:51:40 +0000
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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?