计算（(2+2i）∧4）/((1-3½)∧5)=
展开全部
((2+2i)^4)/((1- √3)^5)
z=2+2i
|z|=2√2
argz =arctan(2/2) = π/4
z=2√2(cos(π/4) +isin(π/4))
z^4 = [2√2( cos(π/4) +isin(π/4) )]^4
= 64( cosπ +isinπ )
=-64i
(1- √3)^5
=1-5√3+10(√3)^2-10(√3)^3+5(√3)^4 -(√3)^5
=(1+30+45)+(-5-30-9)√3
=76-44√3
((2+2i)^4)/((1- √3)^5)
=64i/(44√3-76)
=64(44√3+76)i/32
=2(44√3+76)i
z=2+2i
|z|=2√2
argz =arctan(2/2) = π/4
z=2√2(cos(π/4) +isin(π/4))
z^4 = [2√2( cos(π/4) +isin(π/4) )]^4
= 64( cosπ +isinπ )
=-64i
(1- √3)^5
=1-5√3+10(√3)^2-10(√3)^3+5(√3)^4 -(√3)^5
=(1+30+45)+(-5-30-9)√3
=76-44√3
((2+2i)^4)/((1- √3)^5)
=64i/(44√3-76)
=64(44√3+76)i/32
=2(44√3+76)i
本回答由网友推荐
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
