|z-1|=√[(x-1)^2+y^2]=1
所以
(x-1)^2+y^2=1
设x=sint+1 y=cost
则
|z|=√[x^2+y^2]
=√[(sint+1)^2+cos^2t]
=√(sin^2t+2sint+1+cos^2t)
=√(2sint+2)
因为-1