计算:(1-√3i)(cosx+isinx)÷(1-i)(cosx-isinx),也是一个复数的问题.求解答QAQ

看那华灯初上时 1年前 已收到1个回答 举报

520didicat 幼苗

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[(1-√3i)(cosx+isinx)]÷(1-i)(cosx-isinx)
=[2(1/2-√3/2*i)(cosx+isinx)]÷[√2(√2/2-√2/2*i)][cos(-x)+isin(-x)]
=[2(cos5π/3+isin5π/3)*(cosx+isinx)]÷[√2(cos7π/4+isin7π/4)]*[(cos(-x)+isin(-x))]
=√2[cos(5π/3-7π/4+x-x)+isin(5π/3-7π/4+x-x)]
=√2[cos(-π/12)+sin(-π/12)]
=√2*[(√6+√2)/4-(√6-√2)/4*i]
=(√3+1)/2-(√3-1)/2*i

1年前

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