Re: Division of two complex numbers

On 1/20/2025 2:00 PM, Chris M. Thomasson wrote:

On 1/20/2025 11:29 AM, Moebius wrote:

Am 20.01.2025 um 20:21 schrieb Richard Hachel:

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

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

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No. You can't.

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Yes, we can. :-)
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You don't know what i is but

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We know that i is i, i e C and that i² = -1, idiot.
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If it is  1 then  1² = 1.
If it is -1 then -1² = 1

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Yeah, but since i ISN'T 1 or -1, this is immaterial. Go that, idiot?

 Ahhh, this would be a problem with Hachel's thought process on the nature of i.

For some reason it makes me think about the n-ary roots of a complex number. Take (1+0i). Wrt 2-ary, it has two unique roots.
r0 = -1+0i
r1 = 1+0i
r0^2 = 1+0i
r1^2 = 1+0i
They are really fun to play with and they are very useful. Did some work where I store data in the roots.

 

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Again, i is i, i e C and i² = -1.

  

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<facepalm>
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