设z=x+yi, (z+2)/(z-2)=ci故z+2=(z-2)*cix+yi+2=(x+yi-2)*ci=c(x-2)i-yc即x+2=-yc,y=c(x-2)x+2=-y*y/(x-2)x^2-4=-y^2x^2+y^2=4z+a-3i=x+yi+a-3i=(x+a)+(y-3)i|z+a-3i|=根号[(x+a)^2+(y-3)^2]=...你题目中怎么会又...enen ，谢谢了