复数计算
(√3i-1)^10+(-√3i-1)^10
人气:186 ℃ 时间:2020-05-19 03:33:19
解答
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)+isi...
推荐
猜你喜欢