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
![复数计算题(1-√3i)^15/(1-i)^30+(√3+i)^15/(1+i)^20](/uploads/image/z/361796-68-6.jpg?t=%E5%A4%8D%E6%95%B0%E8%AE%A1%E7%AE%97%E9%A2%98%EF%BC%881-%E2%88%9A3i%EF%BC%89%5E15%2F%281-i%29%5E30%2B%28%E2%88%9A3%2Bi%29%5E15%2F%281%2Bi%29%5E20)
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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