Re: New equation

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

On 03/01/2025 11:11 AM, Jim Burns wrote:

On 2/28/2025 6:08 PM, Ross Finlayson wrote:

On 02/28/2025 10:46 AM, Jim Burns wrote:

The complex field does not fail at being a field.

>

Oh, I read a definition of complex numbers
and point out that division is non-unique.

>
(c+d𝑖)/(a+b𝑖) = (c+d𝑖)⋅(a+b𝑖)⁻¹
>
x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
is unique.
>
(x+y𝑖) = (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)
>
⎛ (a+b𝑖)⋅(x+y𝑖) = (ax-by)+(bx+ay)𝑖 = 1
⎜
⎜ ax-by = 1
⎜ bx+ay = 0
⎜
⎜ a²x-aby = a
⎜ b²x+aby = 0
⎜ (a²+b²)x = a
⎜ x = a/(a²+b²)
⎜
⎜ abx-b²y = b
⎜ abx+a²y = 0
⎜ (a²+b²)y = -b
⎜ y = -b/(a²+b²)
⎜
⎜ x+y𝑖 = (a-b𝑖)/(a²+b²)
⎜
⎜ (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)
⎝ uniquely, for a²+b² ≠ 0

>
Not the other way around though

What do you mean by that?

Oh, I read a definition of complex numbers
and point out that division is non-unique.

Non.zero complex division is uniquely.valued.
(c+d𝑖)/(a+b𝑖) = x+y𝑖
c+d𝑖 = (a+b𝑖)⋅(x+y𝑖)
c+d𝑖 = (ax-by)+(bx+ay)𝑖
ax-by = c
bx+ay = d
a²x-aby = ac
b²x+aby = bd
(a²+b²)x = ac+bd
x = (ac+bd)/(a²+b²)
abx-b²y = bc
abx+a²y = ad
(a²+b²)y = ad-bc
y = (ad-bc)/(a²+b²)
(c+d𝑖)/(a+b𝑖) = x+y𝑖 =
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
Do you (RF) consider
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
multi.valued?
If so, why?