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求复数z=tanθ-i(π/2
题目详情
求复数z=tanθ-i(π/2
▼优质解答
答案和解析
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))
=-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))
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