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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: Equation complexe
Date: Tue, 25 Feb 2025 15:31:08 -0800
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On 2/25/2025 2:09 PM, Richard Hachel wrote:
> Le 25/02/2025 à 22:49, "Chris M. Thomasson" a écrit :
>> On 2/25/2025 6:23 AM, Richard Hachel wrote:
>>> x^4=-81
>>>
>>> What is x?
>>
>>
>> Try to forgive the floating point precision aspects, but, the roots 
>> are the r's, raising them to the 4'th power gives the p's:
>> _____________________
>> r0 = (2.12132,2.12132)
>> r1 = (-2.12132,2.12132)
>> r2 = (-2.12132,-2.12132)
>> r3 = (2.12132,-2.12132)
>>
>> p0 = (-81,-7.08124e-06)
>> p1 = (-81,-1.93183e-06)
>> p2 = (-81,-7.53158e-05)
>> p3 = (-81,4.57051e-05)
>> _____________________
>>
>>
>> To gain a root, here is my code:
>> _____________________
>> ct_complex
>> root_calc(
>>      ct_complex const& z,
>>      int p,
>>      int n
>> ) {
>>      float radius = std::pow(std::abs(z), 1.0 / p);
>>      float angle_base = std::arg(z) / p;
>>      float angle_step = (CT_PI * 2.0) / p;
>>      float angle = angle = angle_step * n;
>>
>>      ct_complex c = {
>>          std::cos(angle_base + angle) * radius,
>>          std::sin(angle_base + angle) * radius
>>      };
>>
>>      return c;
>> }
>> _____________________
>>
>>
>> Also, this is not using floating point for roots, just signed integers.
> 
> This is quite complicated, 

Actually, it's quite straightforward wrt getting at the roots for a 
target number. Even using signed integers for the power.


> where I propose to use the nature of the 
> imaginary number i in a somewhat particular way, and according to the 
> new idea that i is not only defined by i²=-1 or i=sqrt(-1), but rather 
> with the generalized idea that for all x, i^x=-1.
> A bit like if this imaginary was the antithesis of 1 where for all x, 
> then 1^x=1.
> 
> With this technique, we immediately have x=3i.
> 
> R.H.