若复数z满足|z+1|=2|z-1| 试判断 复数z在复平面上对应点的轨迹图形 并求使｜z｜最大时的

[(x+1)^2 + y^2]^0.5 = 2* [(x-1)^2 + y^2]^0.5
(x+1)^2 + y^2 = 4* [(x-1)^2 + y^2]
3y^2 +(3x-1)(x-3) = 0
3y^2 + 3(x - (5/3))^2 = 16/3
(x - (5/3))^2 + y^2 = (4/3)^2
所以轨迹是圆,圆心在(5/3,0),半径为(4/3)
|Z| = (x^2+y^2)^0.5
前述圆形式展开得
x^2+y^2 = (4/3)^2-(5/3)^2 + (10/3)*x
右式最大x存在于圆上点(3,0),
Z = 3 + 0i
得 x^2+y^2
= (4/3)^2-(5/3)^2 + (10/3)*3 = 9
因此最大|Z|
= 9^0.5 = 3