Re: Division of two complex numbers

Le 20/01/2025 à 20:29, Moebius a écrit :

Am 20.01.2025 um 20:21 schrieb Richard Hachel:

Le 20/01/2025 à 19:33, Jim Burns a écrit :

 

((a + ib)(c + id))⃰ =
(ac + i(ad+bc) + i²bd)⃰ =

 No. You can't.

 Yes, we can. :-)
 

You don't know what i is but

 We know that i is i, i e C and that i² = -1, idiot.
 

If it is  1 then  1² = 1.
If it is -1 then -1² = 1

 Yeah, but since i ISN'T 1 or -1, this is immaterial. Go that, idiot?
 Again, i is i, i e C and i² = -1.
 <facepalm>

 No.
 i²=-1 ----> (i)(i)=-1
This has no solution in reality, because we cannot, for example, be here and there, or have two different ages at the same time.
We will then consider that another factor will intervene, for example time. How many students are there in Mrs. Martin's class, 25 or 9? Both at the same time, but not at the same moment.
We then set z=16+9i where i is both 1 and -1.
In the morning, it is 1, in the evening (catch-up class for adults), it is -1.
Now, a mathematical error will occur (in my opinion, but I could be wrong).
When you have to choose, you will have to choose between morning and evening, to know the number of students, and so we will put 1 for morning, and option -1 for evening.
SIMPLY, once the option is chosen, you can no longer say that 1=-1 in the same equation. You start with 1, you stay with 1; you start with -1, you stay with -1.
Now, a mathematical error will occur (in my opinion, but I could be wrong).
When you have to choose, you will have to choose between morning and evening, to know the number of students, and so we will put 1 for morning, and option -1 for evening.
SIMPLY, once the option is chosen, you can no longer say that 1 = -1 in the same equation. You start with 1, you stay with 1; you start with -1, you stay with -1.
And you no longer have the mathematical right to say that (i)(i) = (1)(-1) = -1.
Your two options are now only (1)² = 1 and (-1)² = 1
We then arrive at a logical equation that can eliminate i², since it is the same thing on both sides that we will choose.
z1.z2=aa'+bb'+i(ab'+a'b)
where aa'+bb' is the real part and ab'+a'b the imaginary part.
In this regard, let's take z1=1+i and z2=2-2i
What becomes of Z=z1+z2?
What becomes of Z=z1*z2 (in usual mode and in Hachel mode)?
z1.z2=aa'-bb'+i(ab'+a'b) (classic)
Or, z1.z2=aa'+bb'+i(ab'+a'b)    (Hachel)
R.H.