(-2+i)/(1+2i)
=(-2+i)(1-2i)/(1+2i)(1-2i)
=(-2+4i+i+2)/(1²+2²)
=5i/5
=i
=cos(π/2)+isin(π/2)
³√[cos(π/2)+isin(π/2)]
=cos[(π/2+2kπ)/3]+isin[(π/2+2kπ)/3],k=0,1,2
cos(π/6)+isin(π/6)=√3/2+i/2
cos(5π/6)+isin(5π/6)=-√3/2+i/2
cos(3π/2)+isin(3π/2)=-i
(-2+i)/(1+2i)的三个立方根是√3/2+i/2,-√3/2+i/2,-i
复数开方法则
复数r(cosθ+isinθ)的n次方根
n^√r•{cos[(θ+2kπ)/n]+isin[(θ+2kπ)/n],k=0,1,2,...,n-1