利用复数的指数表示计算（-2+i/1+2i)的1/3次?
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(-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

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