求z=1-√3i的三次方根
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z = 1 -√3i
|z| = 2
argz= arctan(-√3/1)
=5π/3
z =2[cos(2kπ+5π/3) + isin(2kπ+5π/3) ]
z^(1/3) = 2^(1/3) .[cos(2kπ/3+5π/9) + isin(2kπ/3+5π/9) ]
k=0
z^(1/3) = 2^(1/3) .[cos(5π/9) + isin(5π/9) ]
k=1
z^(1/3) = 2^(1/3) .[cos(11π/9) + isin(11π/9) ]
k=2
z^(1/3) = 2^(1/3) .[cos(17π/9) + isin(17π/9) ]
|z| = 2
argz= arctan(-√3/1)
=5π/3
z =2[cos(2kπ+5π/3) + isin(2kπ+5π/3) ]
z^(1/3) = 2^(1/3) .[cos(2kπ/3+5π/9) + isin(2kπ/3+5π/9) ]
k=0
z^(1/3) = 2^(1/3) .[cos(5π/9) + isin(5π/9) ]
k=1
z^(1/3) = 2^(1/3) .[cos(11π/9) + isin(11π/9) ]
k=2
z^(1/3) = 2^(1/3) .[cos(17π/9) + isin(17π/9) ]
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