Question

z=(-1+cosθ)+(2+sinθ)i,求z模的最大值以及z模的最小值

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...