Question

复变函数中z=1+cosθ+i sinθ是怎么化简的,

Answer 1

|z|=√[(-1+cosθ)^2+(2+sinθ)^2]

=√[(1-2cosθ+(cosθ)^2)+(4+4sinθ+(sinθ)^2)]

=√(6+4sinθ-2cosθ)

=√[6+√20*(4/√20*sinθ-2/√20*cosθ)]

=√[6+2√5*(2/√5*sinθ-1/√5*cosθ)]

令cosα=2/√5,sinα=1/√5

则原式=√[6+2√5*(sinθcosα-cosθsinα)]

=√[6+2√5*sin(θ-α)]

因为-1