复数计算题(1-√3i)^15/(1-i)^30+(√3+i)^15/(1+i)^20
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复数计算题
(1-√3i)^15/(1-i)^30+(√3+i)^15/(1+i)^20
(1-√3i)^15/(1-i)^30+(√3+i)^15/(1+i)^20
原式=(1-√3i)^15/(-2i)^15+(√3+i)^15/(2i)^10
=[(1-√3i)/(-2i)]^15+(√3+i)^15*2^5/[2^5*2^10*(i)^(8+2)]
=(√3+i)^15/(2)^15+(√3+i)^15*32/[-2^15]
=-31(√3+i)^15/(2)^15
=-31[cosπ/6+isinπ/6)^15
=-31[cos15π/6+isin15π/6)
=-31(cos5π/2+isin5π/2)
=-31(cosπ/2+isinπ/2)
=-31i.
=[(1-√3i)/(-2i)]^15+(√3+i)^15*2^5/[2^5*2^10*(i)^(8+2)]
=(√3+i)^15/(2)^15+(√3+i)^15*32/[-2^15]
=-31(√3+i)^15/(2)^15
=-31[cosπ/6+isinπ/6)^15
=-31[cos15π/6+isin15π/6)
=-31(cos5π/2+isin5π/2)
=-31(cosπ/2+isinπ/2)
=-31i.
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