复变函数1.9的第三和第四小题怎么做？急，谢谢！
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[ (1-√3i)/2]^3
=(1/8) [ 1 - 3(√3i) + 3(√3i)^2 -(√3i)^3 ]
=(1/8) ( -8 )
=-1
(4)
z
= -2+2i
=2√2 ( -1/√2 + i/√2)
=2√2 [ cos(3π/4) + isin(3π/4)]
(-2+2i)^(1/4)
=z^(1/4)
=2^(3/8) [ cos(3π/16) + isin(3π/16)]
=(1/8) [ 1 - 3(√3i) + 3(√3i)^2 -(√3i)^3 ]
=(1/8) ( -8 )
=-1
(4)
z
= -2+2i
=2√2 ( -1/√2 + i/√2)
=2√2 [ cos(3π/4) + isin(3π/4)]
(-2+2i)^(1/4)
=z^(1/4)
=2^(3/8) [ cos(3π/16) + isin(3π/16)]
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