问题: 请教一个数学题 急！急！！
希望大家耐心做。麻烦大家把过程写详细一些。谢谢。
解答:
1-z+z^2=0，由求根公式有：z=(1±√3i)/2
即，z=cos(π/3)±isin(π/3)
那么，当z=cos(π/3)+isin(π/3)时：
(1+z^1000)/z^2000
=[1+cos(1000π/3)+isin(1000π/3)]/[cos(2000π/3)+isin(2000π/3)]
=[1+cos(4π/3)+isin(4π/3)]/[cos(2π/3)+isin(2π/3)]
=[1-(1/2)-(√3/2)i]/[(-1/2)+(√3/2)i]
=[(1/2)-(√3/2)i]/[(-1/2)+(√3/2)i]
=-1
当z=cos(π/3)-isin(π/3)时：
(1+z^1000)/z^2000
=[1+cos(1000π/3)-isin(1000π/3)]/[cos(2000π/3)-isin(2000π/3)]
=[1+cos(4π/3)-isin(4π/3)]/[cos(2π/3)-isin(2π/3)]
=[1-(1/2)+(√3/2)i]/[(-1/2)-(√3/2)i]
=[(1/2)+(√3/2)i]/[(-1/2)-(√3/2)i]
=-1
综上，原式=-1
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