(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/04/30 18:56:51

(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8
√3是根号3

-√3/2-1/2i=cos(-2π/3)+isin(-2π/3)
(2+2i)/(1-√3i)=2√2(cosπ/4+isinπ/4)/2[cos(-π/3)+isin(-π/3)]
所以(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8
=[cos(-2π/3)+isin(-2π/3)]^12+[2√2(cosπ/4+isinπ/4)/2[cos(-π/3)+isin(-π/3)]]^8
=e^-8πi+16e^14πi/3=1+16e^2πi/3=-8√3+16+8i

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