Question

ʕ •ᴥ•ʔ

設ω=cos(2π/5)+isin(2π/5)
求1/(1+ω)+1/(1+ω²)+1/(1+ω³)+1/(1+ω⁴)=？

Answer 1

✨ 最佳解答 ✨

anonymous_user

ω⁵=1
所以1/(1+ω)+1/(1+ω²)+1/(1+ω³)+1/(1+ω⁴)
=[ 1/(1+ω)+1/(1+ω⁴)]+[1/(1+ω²)+1/(1+ω³)]
=[ 1/(1+ω)+1/(1+(1/ω))]+1/(1+ω²) +1/(1+(1/ω²))
= [1/(1+ω)+ω/(1+ω)]+[1/(1+ω²)+ω²/(1+ω²)]
=1+1=2

ʕ •ᴥ•ʔ

看懂了 感謝您

Answer 2

我就是我

如圖

ʕ •ᴥ•ʔ

為什麼ω=-ω⁴, ω²=-ω³?

我就是我

你畫圖看看

ʕ •ᴥ•ʔ

這樣不是共軛複數嗎