设z=a+bi则z^2=a^2-b^2+2ab *i所以a^2-b^2=82ab=6得出a=3 b=1 或a=-3 b=-1z=3+i或-3-i原式=（8+6i)(3+i)-16(3+i)-100/(3+i)=(24-6+26i)-(48+16i)-100(3-i)/[(3+i)(3-i)]=-30+10i -10(3-i)=-60+20i当a=-3 b=-1时原式=...