Question

设z=cosθ+isinθ(0<π<2π),证明(1+z)/(1-z)=icotθ/2

Answer 1

(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

Answer 2

(1+z)/(1-z)=((1+cosθ+isinθ)(1-cosθ+isinθ))/((1-cosθ-isinθ)(1-cosθ+isinθ))=2isinθ/(2-2cosθ)=icot(θ/2)