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On 3/10/2025 1:33 AM, efji wrote:Yes !Le 10/03/2025 à 03:03, Chris M. Thomasson a écrit :Humm... I am not all that familiar with the split complex numbers. Something like this?
>>>
Indeed. Sorry for the stupid question, but the following parts of RH's description:
>
Z=aa'+bb'+i(ab'+a'b) and not Z=aa'-bb'+i(ab'+a'b).
>
Means a = x component, b = y component, right? ;^o
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
>
But forget the pathetic egotic and just try a+jb=(a,b) un R^2 with the rule (a+ib)*(a'+jb') = aa'+bb'+j(ab'+a'b) which is the rule on the split-complex set, (thus j^2=1).
>
glm::vec2
ct_complex_split_mul(
const glm::vec2& z1,
const glm::vec2& z2
) {
return {
z1.x * z2.x + z1.y * z2.y,
z1.x * z2.y + z1.y * z2.x
};
}
Where:
glm::vec2 s0 = { 0, 1 };
std::cout << "s0 * s0 = " << ct_complex_split_mul(s0, s0) << "\n";
outputs:
s0 * s0 = (1, 0)
?
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