Question

1-cosθ+isinθ(0≤θ≤π) 写出该复数的三角式和指数形式

答案是2sin(θ/2)(cos((π- θ）/2)+isin((π- θ）/2))=2sin(θ/2)e^i((π- θ）/2)

求过程啊~~

Answer 1

利用cosθ=1-2(sinθ)^2 和sinθ=2sin(θ/2)cos(θ/2), 1-cosθ+isinθ=2sin(θ/2)(sin(θ/2)+icos(θ/2))=
2sin(θ/2)(cos((π- θ）/2)+isin((π- θ）/2))=2sin(θ/2)e^i((π- θ）/2)