设z=a+ib;z+(10/z)为实数,则表明虚部为0,即：ib-i10b/(a^2+b^2)=0；得到：b-10b/(a^2+b^2)=0若b=0,由不等式范围可得a=3；若b!=0,则a^2+b^2=10,即a=3、b=4或b=3、a=4;考虑z+(10/z)的实部为a+10a/(a^2+b^2),将a^2+b^2=...