一道关于复数的题

设Z=x+yi,(x,y∈R),则Z+2/Z=x+yi+2/(x+yi)=x+2x/(x²+y²)+[y-2y/(x²+y²)]i
由Z+(2/Z) ∈R得y-2y/(x²+y²)=0.
所以y=0或x²+y²=2
若y=0,则z=x.|(Z+i)/(Z-i)|=|(x+i)/(x-i)|=|x+i|/|x-i|=1恒成立.
若x²+y²=2,则|(Z+i)/(Z-i)|=|x+(y+1)i|/|x+(y-1)i|=√[x²+(y+1)²]/√[x²+(y-1)²]=√(2y+3)/√(-2y+3)=1,得到y=0.
故z是任意实数.