设z=x+yi(x,y∈R),
则
.
z=x−yi,
代入z+
.
z+z•
.
z=0,得:
x+yi+x−yi+(
x2+y2)2=0,
即x2+y2+2x=0.
整理得:(x+1)2+y2=1.
∴复数z在复平面内对应的点的轨迹是圆.
故选:A.
设z=x+yi(x,y∈R),
则
.
z=x−yi,
代入z+
.
z+z•
.
z=0,得:
x+yi+x−yi+(
x2+y2)2=0,
即x2+y2+2x=0.
整理得:(x+1)2+y2=1.
∴复数z在复平面内对应的点的轨迹是圆.
故选:A.