1z=a+bi,z+2/z为实数a+bi+2/(a+bi)=a+bi+[2/(a^2+b^2)](a-bi)b-2b/(a^2+b^2)=0a^2+b^2=2|z|=√22z=a+√(2-a^2)i(z+1)/(z-1)=[(a+1)+√(2-a^2)i]/[(a-1)+√(2-a^2)i]=[(a+1)+√2-a^2)i][(a-1)-√(2-a^2)i]/[(a-1)^2+(...