已知复数Z=2倍根号3-2i,将Z化为三角形式;计算Z的6次方
人气:200 ℃ 时间:2019-10-23 11:46:46
解答
z=4(√3/2-i/2)
=4(cos11π/6+isin11π/6)
z^6
=4^6(cos11π+isin11π)
=-4096������ʦ�����ǵĴ������ʽ��Z=4[cos(-��/6)+isin(-��/6)]һ�����ɰ�
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