Sujet : Re: New equation
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 28. Feb 2025, 14:50:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <fce8aaae-4ac3-4909-99df-60491fc4de7e@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
User-Agent : Mozilla Thunderbird
On 2/27/2025 11:51 PM, Ross Finlayson wrote:
On 02/27/2025 01:46 AM, efji wrote:
Le 27/02/2025 à 05:19, Ross Finlayson a écrit :
Division in complex numbers is opinionated,
not unique.
:)
Hachel has a brother !
A BS-philosophical version of Hachel.
Let's park them together.
Division in complex numbers most surely is
non-unique, [...]
In the complex field,
division is unique, except by 0,
and division by 0 isn't at all.
Because the complex field is a field.
A field has operations '+' '⋅' such that
both are associative and commutative,
have identities 0 1 and inverses -x x⁻¹,
except 0⁻¹, and '⋅' distributes over '+'
Consider inverse(s?) x⁻¹ x⁻¹′
x⁻¹⋅x⋅x⁻¹′ = x⁻¹⋅x⋅x⁻¹′
x⁻¹⋅1 = 1⋅x⁻¹′
x⁻¹ = x⁻¹′
The complex field is a field,
in part, because 𝑖² = -1
from which it must follow that,
for (a+b𝑖)⁻¹ = x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
⎛
⎜ (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
It's possible that
you are confusing division with
logarithm or square root or some such.
Furthermore, if you don't know usual derivations
of Fourier-style analysis and for example about
that the small-angle approximation is a linearisation
and is an approximation and is after a numerical method,
you do _not_ know.
Then about integral analysis and this sort of
"original analysis" and about the identity line
being the envelope of these very usual integral
equations, it certainly is so.
Technobabble.
And not in a good way.
[...]