设z=x+iy
│z-4│=│z-4i│==>x=y
Im(z+(14-z)/(z-1))=0 (z=x+jy=x+jx)
==> 2x(x-3)(x+2)/(2x^2-2x+1)=0
==>x=0或3或-2
==>z=0或3+i3或-2-i2