设z=x+yi
丨z+1丨=√[(x+1)^2+y^2]
丨z+i丨=√[x^2+(y+1)^2]
丨z+1丨²-丨z+i丨²=x^2+2x+1+y^2-x^2-y^2-2y-1=1
2x=2y=1
在复平面上满足丨z+1丨²-丨z+i丨²=1的复数z对应的点z轨迹是直线