（1）由题意知,z=x(1+i)-y(2+i) =x+xi-2y-yi =(x-2y)+(x-y)i因为复数z是纯虚数所以x-2y=0,x-y≠0,即x=2y且x≠y,另外由|z|=1可得,(x-2y)^2+(x-y)^2=1展开得2x^2-6xy+5y^2=1代入x=2y得8y^2-12y^2+5y^2=1得y=1或-1当y=1...