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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: Division of two complex numbers
Date: Mon, 20 Jan 2025 12:44:36 -0800
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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);
}