(a) Why i^2=-1?
b^2=-1?
c^2=-1?
(b)
(a+bi)(c+di)=(ac-bd)(ad+bc)i
why not =ac+adi+cbi+bdi^2
=a(c+di)+bi(c+di)
=(a+bi)(c+di)
(c)a+bi/c+di why =ac+bd/c^2+d^2 +bc-ad/c^2=d^2 (i)
Complex number
2012-03-10 11:20 pm
回答 (2)
2012-03-11 2:20 am
✔ 最佳答案
a. definition, 問題同點解pi係圓周率一樣b. ac+adi+cbi+bdi^2=ac+adi+cbi-bd=(ac-bd)+(ad+bc)i
你睇漏左個加
c. (a+bi)/(c+di)
=(a+bi)(c-di)/(c+di)(c-di)
=[(ac-bd)+(ad+bc)i]/(c^2+d^2)
2012-03-12 6:01 am
(a) i^2 = [sqrt (-1)]^2 = -1
i = sqrt (-1) is the definition, but it does not necessarily imply that i^2 = -1 is also the definition.
i = sqrt (-1) is the definition, but it does not necessarily imply that i^2 = -1 is also the definition.
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