Re: Division of two complex numbers

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

Division of two complex numbers.
>
Now let's set Z=(a+ib)/(a'+ib')
with
z1=a+ib
and
z2=a'+ib'
>
What becomes of Z=A+iB?

x + iy  =  (1 + i0)/(c + id)
(c + id)(x + iy)  =  1 + i0
(cx + i²dy) + (cy + dx)i  =  1 + i0
cx + i²dy  =  1
dx + cy  =  0
c²x + i²cdy  =  c
i²d²x + i²cdy  =  0
(c²-i²d²)x  =  c
cdx + i²d²y  =  d
cdx + c²y  =  0
(c²-i²d²)y  =  -d
x + iy  =  (1 + i0)/(c + id)
x + iy  =  (c - id)/(c²-i²d²)
If i² ≠ 0 then
then some c + id  ≠  0 + i0
do not have inverses.
Above, I have asserted that i² is real (is east.west).
Real i² can be proven from the assumption (w⋅z)⃰ = w⃰⋅z⃰
[1]
There might be yet another argument showing that
i² is specifically -1
Or there might not be. I'm not sure right now.
Even if there isn't an argument for i²=-1,
the north.south units can be re.defined
so that it is so.
That leaves us with essentially
only one complex multiplication.
[1]
(w⋅z)⃰ = w⃰⋅z⃰
w = a + ib
z = c + id
i² = sₑ + i⋅sₙ
(w⋅z)⃰  =
((a + ib)(c + id))⃰  =
(ac + i(ad+bc) + i²bd)⃰  =
(ac + i(ad+bc) + (sₑ+i⋅sₙ)bd)⃰  =
((ac+sₑbd) + i(ad+bc+sₙbd))⃰  =
(ac+sₑbd) - i(ad+bc+sₙbd)
w⃰⋅z⃰  =
(a + ib)⃰⋅(c + id)⃰  =
(a - ib)⋅(c - id)  =
ac - i(ad+bc) + i²bd  =
ac - i(ad+bc) + (sₑ+i⋅sₙ)bd  =
(ac+sₑbd) - i(ad+bc-sₙbd)
(w⋅z)⃰ - w⃰⋅z⃰  =  -2i⋅sₙbd
(w⋅z)⃰ = w⃰⋅z⃰  ⇒  sₙ = 0
i² = sₑ