复数计算(√3i-1)^10+(-√3i-1)^10
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/04 16:55:32
复数计算
(√3i-1)^10+(-√3i-1)^10
(√3i-1)^10+(-√3i-1)^10
原式=[2^10(-1/2+√3i/2)^10]+2^10[(1/2+√3i/2)^10]
=2^10[cos(2π/3)+isin(2π/3)]^10+2^10[cos(π/3)+isin(π/3)]^10
=1024[cos(20π/3)+isin(20π/3)]+1024[cos(10π/3)+isin(10π/3)]
=1024[cos(2π/3)+isin(2π/3)+1024[cos(4π/3)+isin(4π/3)]
1024(-1/2+√3i/2)+1024(-1/2-√3i/2)
=-1024.
=2^10[cos(2π/3)+isin(2π/3)]^10+2^10[cos(π/3)+isin(π/3)]^10
=1024[cos(20π/3)+isin(20π/3)]+1024[cos(10π/3)+isin(10π/3)]
=1024[cos(2π/3)+isin(2π/3)+1024[cos(4π/3)+isin(4π/3)]
1024(-1/2+√3i/2)+1024(-1/2-√3i/2)
=-1024.
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