设z=cosθ+isinθ(0<π<2π),证明(1+z)/(1-z)=icotθ/2
答案:2 悬赏:20
解决时间 2021-01-16 09:49
- 提问者网友:遮云壑
- 2021-01-15 21:11
设z=cosθ+isinθ(0<π<2π),证明(1+z)/(1-z)=icotθ/2
最佳答案
- 二级知识专家网友:从此江山别
- 2021-01-15 21:17
(1+z)/(1-z)=z
解析:
1±z
=(1+cosθ)±isinθ
=2cos²(θ/2)±i2sin(θ/2)cos(θ/2)
=2cos(θ/2)[cos(θ/2)±isin(θ/2)]
∴
(1+z)/(1-z)
=[cos(θ/2)+isin(θ/2)]/[cos(θ/2)-isin(θ/2)]
=e^(iθ/2)/[e^(-iθ/2)]
=e^(iθ)
=z
解析:
1±z
=(1+cosθ)±isinθ
=2cos²(θ/2)±i2sin(θ/2)cos(θ/2)
=2cos(θ/2)[cos(θ/2)±isin(θ/2)]
∴
(1+z)/(1-z)
=[cos(θ/2)+isin(θ/2)]/[cos(θ/2)-isin(θ/2)]
=e^(iθ/2)/[e^(-iθ/2)]
=e^(iθ)
=z
全部回答
- 1楼网友:舊物识亽
- 2021-01-15 21:54
(1+z)/(1-z)=((1+cosθ+isinθ)(1-cosθ+isinθ))/((1-cosθ-isinθ)(1-cosθ+isinθ))=2isinθ/(2-2cosθ)=icot(θ/2)
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