复数计算:(1/2+(根号3/2i))^5+(-1/2+(根号3/2i))^3
1个回答
展开全部
答：把原式化简成三角形式。 利用 [r(cosθ+isinθ)]^n=（r^n）[cosnθ+isinnθ] 公式计算
[cos（π/3）+sin(π/3)i]^5+[cos(2/3π）+sin(2/3π)i]³
=[cos(5Xπ/3)+sin(5Xπ/3]+[cos(3X2/3π)+sin(3X2/3π)i]
=[cos(2π-π/3)+cos(2π)]+[sin(2π-π/3)+sin(2π)]i
=【cos(π/3)-1】+[-sin(π/3)-0]i
=(1/2-1)-(√3/2)i
=-(1/2+√3/2)i
[cos（π/3）+sin(π/3)i]^5+[cos(2/3π）+sin(2/3π)i]³
=[cos(5Xπ/3)+sin(5Xπ/3]+[cos(3X2/3π)+sin(3X2/3π)i]
=[cos(2π-π/3)+cos(2π)]+[sin(2π-π/3)+sin(2π)]i
=【cos(π/3)-1】+[-sin(π/3)-0]i
=(1/2-1)-(√3/2)i
=-(1/2+√3/2)i
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
