求复数z=tanθ-i(π/2

分类: 作业答案 • 2022-04-05 19:41:38

问题描述：

求复数z=tanθ-i(π/2

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