设复数z满足zz-+(2-i)z+(2+i）z- +4=0 求证 z在复平面上所对应的点到复数-2-i在复平面上所对应的

设z=x+yi
zz-+(2-i)z+(2+i）z- +4=0
x^2+y^2+(2-i)(x+yi)+(2+i)(x-yi)+4=0
x^2+y^2+2x+2yi-xi+y+2x-2yi+xi+y+4=0
x^2+y^2+4x+2y+4=0
(x+2)^2+(y+1)^2=1,即在复平面表示为
|z-(-2-i)|=1
所以原式成立