Re: Division of two complex numbers

Le 21/01/2025 à 11:37, Moebius a écrit :

Am 21.01.2025 um 11:27 schrieb Moebius:>

Let's write the complex number (a, b) in the usual form a + ib. Then
      (a + ib)^2 = a^2 + i2ab + (ib)^2

 Vous allez trop vite.
 (a + ib)^2 = a²+2abi+(ib)(ib)  YOU say (ib)(ib), it's the same thing as i²b² with i²=-1.  With careful reflection, I realize that if we change the definition of i, that is to say if we define it as a unit being at the same time its unit, but also its opposite, its inverse, the inverse of its opposite and the opposite of its inverse, all at the same time, we realize that, not knowing i, and by attributing to it the two values
in the same equation, we have i²=(i1)(i2)=(1)(-1)=-1.
But there are two solutions which are i1 and i2.
If I take i1, I must maintain i1, and the same for i2.
It comes that if the imaginary part remains correct, i(ab'+a'b), it is no longer true of the real part, because whether we take i1 or i2, each time i²=1, and we come across a position simplification of i, and not a negative simplification.
We can then write:
Z=aa'+bb'+i(ab'+a'b) which seems to me to be the correct solution (verified by statistics (example of the students of Plougastel).
The same for the division:
Z=z1/z2
We obtain another mathematical appearance, and
Z=[(aa'-bb')/(a'²-b'²)] + i [(ba'-ab')/(a'²-b'²)]
 

by the binomial formula (!). Hence
      (a + ib)^2 = a^2 + i2ab + (-1)b^2 = (a^2 - b^2) + i2ab .

 No.

 And hence
      (a, b)^2 = (a^2 - b^2) + i2ab  (!)

 No.

 or rather
      (a, b)^2 = (a^2 - b^2, 2ab)
 would be correct.

 Incorrect.
 z1=2+2i
 z2=1-i
 Z=z1*z2
 You say that Z=4, and me Z=0.
 Le produit de deux complexes orthogonaux étant nul chez moi, égal à 4 chez vous.
 R.H.