Re: Division of two complex numbers

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

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

Division of two complex numbers.
>
Now let's set Z=(a+ib)/(a'+ib')

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

No. You can't.
=(ac + i(ad+bc) + (ib*id)⃰  You don't know what i is but at this moment, you know that it is both i=1 and i=-1.
If it is 1 then 1²=1.
If it is -1 then -1²=1
In any case, its square will be 1 because in your two solutions, you will have to give your choice, but not both at the same time.
 

(ac + i(ad+bc) + (sₑ+i⋅sₙ)bd)⃰  =
((ac+sₑbd) + i(ad+bc+sₙbd))⃰  =
(ac+sₑbd) - i(ad+bc+sₙbd)

 ac+bd+i(ad+bc) with i=(+/-)1 at final choise.
R.H.