求解一题复数复数z、w满足zw+2iz-2iw+1=0,求证：若|z|=sqrt(3),则|w-4i|是常数并求出该常数

复数z、w满足zw+2iz-2iw+1=0,
|z|=√3,设z=√3(cost+isint),则
√3w(cost+isint)+2√3(-sint+icost)-2iw+1=0
整理得w[√3cost+i(√3sint-2)]=2√3sint-1-i*2√3cost,
∴w=(2√3sint-1-i*2√3cost)/[√3cost+i(√3sint-2)],
∴w-4i=[2√3sint-1-i*2√3cost-4√3icost+4(√3sint-2)]/[√3cost+i(√3sint-2)]
=[6√3sint-9-i*6√3cost]/[√3cost+i(√3sint-2)]
∴|w-4i|=3|2√3sint-3-2√3icost|/|√3cost+i(√3sint-2)|
=3√[(2√3sint-3)^2+(2√3cost)^2]/√[(√3cost)^2+(√3sint-2)^2]
=3√[(21-12√3sint)/(7-4√3sint)]
=3√3.