z=(-1/2)+(開方3/2)i simplify z^100 有冇高手識做?
回答 (1)
2017-01-24 2:19 pm
Sol
z=(－1/2)+i√3/2=Cos(－π/3)+iSin(－π/3)
z^100= Cos(－100π/3)+iSin(－100π/3)
= Cos(－100π/3+34π)+iSin(－100π/3+34π)
= Cos(2π/3)+iAin(2π/3)
=－1/2+i√3/2
z=(－1/2)+i√3/2=Cos(－π/3)+iSin(－π/3)
z^100= Cos(－100π/3)+iSin(－100π/3)
= Cos(－100π/3+34π)+iSin(－100π/3+34π)
= Cos(2π/3)+iAin(2π/3)
=－1/2+i√3/2
收錄日期: 2021-04-30 22:08:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170124045432AA9q32d