求算法:極式問題
回答 (2)
2012-06-22 11:14 pm
✔ 最佳答案
Sol(√3－i)^10=2^10*(√3/2－i/2)^10=1024*[Cos(330度)+iSin(330度)]^10=1024*[Cos(3300度)+iSin(3300度)]=1024*[Cos(3300度－9*360度)+iSin(3300度－9*360度)]=1024*[Cos(60度)+iSin(60度)]=1024*[(1/2)+i(√3/2)]=512+512√3i
2012-06-23 3:54 am
本題用負角可讓計算量少些
(√3-i)^10
=2^10*(√3/2-i/2)^10
=2^10*[cos(-30度)+isin(-30度)]^10
=2^10*[cos(-300度)+isin(-300度)]
=2^10*[cos(60度)+isin(60度)]
=2^10*[(1/2)+i(√3/2)]
=512+512√3i
(√3-i)^10
=2^10*(√3/2-i/2)^10
=2^10*[cos(-30度)+isin(-30度)]^10
=2^10*[cos(-300度)+isin(-300度)]
=2^10*[cos(60度)+isin(60度)]
=2^10*[(1/2)+i(√3/2)]
=512+512√3i
收錄日期: 2021-04-30 16:39:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120622000010KK03954