复数三角形式、复数次方运算

2025-04-14 05:42:35
推荐回答(1个)
回答1:

z^2=cos(4π/5)+isin(4π/5)=-cos(π/5)+isin(π/5),
z^4=(z^2)²=cos(2π/5)-isin(2π/5)。
z^8=(z^4)²=cos(4π/5)-isin(4π/5)=-cos(π/5)-isin(π/5)。所以z^2·z^8=z^10=1,所以 z^12=z^2。
z^16=(z^8)²=cos(2π/5)+isin(2π/5)。所以原式=1+z^4+z^8+z^2+z^16=
1+cos(2π/5)-isin(2π/5)-cos(π/5)-isin(π/5)-cos(π/5)+isin(π/5)+cos(2π/5)+isin(2π/5)
含i的项全部消掉。所以原式=1+2cos(2π/5)-2cos(π/5)。