Re: Division of two complex numbers

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:
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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:
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z1/z2=[(aa'+bb')/(a'²+b'²)]+i[(ba'-ab')/(a'²+b'²)]
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It was necessary to write:
z1/z2=[(aa'-bb')/(a'²-b'²)]+i[(ba'-ab')/(a'²-b'²)]

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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.

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 That is what I saw.
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 Is not a definition.
 It doesn't explain why.
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We have the same thing with Einstein and relativity.
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[snip unrelated nonsense about your idiotic views on Relativity]

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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.

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I've posted a definition of i (which is NOT i^2 = -1) numerous times. A "positive" definition as you asked for.

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I've already told this idiot:
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Complex numbers can be defined as (ordered) pairs of real numbers.
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Then we may define (in this context):
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          i := (0, 1) .
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 From this we get: i^2 = -1.

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For R.H.
  By the binominal formulas we have: (a, b)^2 = a^2 + 2ab + b^2

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Huh? This is not the binomial formula which is (a + b)^2 = a^2 + 2ab + b^2
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(a, b)^2 does not mean anything without any additional definition/ context.
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  So we get: (0, 1)^2 ) 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1

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you meant  (0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1 ?
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This does not make sense without additional context.
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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 :
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(0, 1) ^ 2 = 0
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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)
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*.