设z=x+yi
|z+3-4i|
=|x+3+(y-4)i|
=√[(x+3)^2+(y-4)^2=2
(x+3)^2+(y-4)^2=4
设x=2sint-3y=2cost+4
则
|z-1|=√(x-1)^2+y^2
=√(2sint-3-1)^2+(2cost+4)^2
=√4sint^2-16sint+16+4cos^2+16cost+16
=√-16sint+16cost+36
=√-16(sint-cost)+36
=√-16√2(sintcos45-costsin45)+36
=√36-16√2sin(t-45)
因为-1<=sin(t-45)<=1
所以√(36-16√2)<=√36-16√2sin(t-45)<=√(36+16√2)
对不上答案?