设z=x+yi
|z-1|=1
(x-1)²+y²=1
设x-1=cosa ,y=sina
|z-i|
=√[x²+(y-1)²]
=√[(cosa+1)²+(sina-1)²]
=√(cos²a+2cosa+1+sin²a-2sina+1)
=√[3+2(cosa-sina)]
=√[3+2√2(cos(a+π/4)]
所以|z-i|的最小值=√(3-2√2)=√2-1
最大值=√（3+2√2)=√2+1