Re: Division of two complex numbers

On 1/20/2025 2:21 PM, Richard Hachel wrote:

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

On 1/20/2025 6:02 AM, Richard Hachel wrote:

Division of two complex numbers.
>
Now let's set Z=(a+ib)/(a'+ib')

>

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

>
No. You can't.

If planar (complex) multiplication has
certain reasonable.looking properties,
it is a theorem that i² has certain properties.
If (w⋅z)* = w*⋅z*
then, for the north (imaginary) unit i
i² is on the east,west (real) axis, i²=(i²)*
If, for each non.0 z,
there is an inverse w = z⁻¹, z⋅w = 1
then i²<0
Those aren't all the reasonable.looking properties
which planar multiplication should have.
If, for each planar pair v,w
there is exactly one two.dimensional z = v⋅w
then your Schrödinger's cat never exists,
not even in a Schrödinger's.cat semi.existent way.

=(ac + i(ad+bc) + (ib*id)⃰
You don't know what i is
but at this moment,
you know that it is both i=1 and i=-1.
If it is 1 then 1²=1.
If it is -1 then -1²=1
In any case, its square will be 1
because in your two solutions,
you will have to give your choice,
but not both at the same time.

Reasoning from incomplete information is
just another day in Mathematics.
How to say i=1 or i=-1 and not say which
is  i ∈ {1,-1}
Planar "multiplication" with i ∈ {1,-1}
is not the multiplication we are looking for.