F.4 Maths
回答 (3)
2011-12-22 7:27 am
✔ 最佳答案
[(1+i)⁹]/[(1-i)⁸]=[(1+i)/(1-i)]⁸(1+i)={[(1+i)(1+i)]/[(1-i)(1+i)]}⁸(1+i)=[(1+2i+i²)/(1-i²)]⁸(1+i)=[(1+2i-1)/(1+1)]⁸(1+i)=[(2i)/2]⁸(1+i)=i⁸(1+i)=1(1+i) =1+i
參考: Hope I Can Help You ! ^_^ ( From Me )
2011-12-20 4:39 am
(1+i)^(9)/(1-i)^(8)
={(1+i)/(1-i)]^8(1+i)
={[(1+i)/(1-i)]*[(1+i)/(1+i)]}^8(1+i)
=[(1+2i-1)/2]^8(1+i)
=(2i/2)^8(1+i)
=i^8(1+i)
=1(1+i)
=1+i
={(1+i)/(1-i)]^8(1+i)
={[(1+i)/(1-i)]*[(1+i)/(1+i)]}^8(1+i)
=[(1+2i-1)/2]^8(1+i)
=(2i/2)^8(1+i)
=i^8(1+i)
=1(1+i)
=1+i
2011-12-20 2:24 am
(1+ i)^9/(1 - i)^8
= (1+ i)^9(1 + i)^8/(1 - i)^8(1 + i)^8
= (1+ i)^17/256
= (√2)^17[cos(17π/4) + isin(17π/4)]/256
= √2[cos(π/4) + isin(π/4)]
= 1 + i
= (1+ i)^9(1 + i)^8/(1 - i)^8(1 + i)^8
= (1+ i)^17/256
= (√2)^17[cos(17π/4) + isin(17π/4)]/256
= √2[cos(π/4) + isin(π/4)]
= 1 + i
收錄日期: 2021-04-11 18:51:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111219000051KK00364