z=(-1+cosθ)+(2+sinθ)i,求z模的最大值以及z模的最小值

|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<=sin(θ-α)<=1
所以√(6-2√5)<=sin(θ+α)<=√(6+2√5)
z模最大值为√(6+2√5)，最小值为√(6-2√5)

Max值=(根号5)+1
Min值＝（根号5）－1