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求复数z=tanθ-i(π/2
人气:342 ℃ 时间:2020-06-22 15:52:33
解答
z=tanθ-i
=-cot(π/2+θ)-i
=-1/sin(π/2+θ)(cos(π/2+θ)+isin(π/2+θ))
=1/sin(π/2+θ)(cos(π+π/2+θ)+isin(π+π/2+θ))
=1/sin(π/2+θ)(cos(θ-π/2)+isin(θ-π/2))π/2<θ<π有用吗?有用,要保证:θ-π/2在0到2π之间。
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