解下列复数方程 1.z^3=-1+i根3 2.z^3=-1

1.z^3=2(cos2π/3+isin2π/3)
因此z=2^(1/3)[cos(2kπ/3+2π/9)+isin(2kπ/3+2π/9)],k=0,1,2
2.z^3=-1=cosπ+isinπ
z=cos(2k+1)π/3+isin(2k+1)π/3,k=0,1,2
z1=cosπ/3+isinπ/3=1/2+i√3/2
z2=cosπ+isinπ=-1
z3=cos5π/3+isin5π/3=1/2-i√3/2