Re: Division of two complex numbers

Am 21.01.2025 um 00:08 schrieb Tom Bola:

Am 20.01.2025 23:26:19 Moebius schrieb:

Am 20.01.2025 um 23:21 schrieb Tom Bola:
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The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about
(0, 1)^2 or (a, b)^2.

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Sure, but remember the definition of a complex number: Z = a+bi, i^2 = -1.
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See?

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See WHAT?!
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Was wir hier brauchen, ist die Definition der MULTIPLIKATION (*) auf C =
{(x, y) : x,y e IR}.
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DARAUS folgt dann
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        i * i = (0, 1) * (0, 1) = ... = (-1, 0)
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See?

 Ja! This is fully straight plus absolutely correct.
 I was just looking for some idea. Aside from that please have a look here:
 (english readers just look for "binomisch")
 https://www.math.kit.edu/iana2/~mandel/seite/schnupperkurs/media/komplex.pdf

Gutes pdf.
Das steht:
"Der Division liegt die _dritte Binomische Formel_ (a + b)(a − b) = a^2 - b^2 zugrunde."
Nicht: (a + b)^2 = a^2 + 2ab + b^2
Got it?
Und es geht dabei um die Division, nicht um (a, b)^2. (*seufz*)
Hint: You wrote (the nonsense): "By the binominal formulas we have: (a, b)^2 = a^2 + 2ab + b^2"
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