已知z=a+bi是虚数,且z+1/z是实数,求证:a^2+b^2=1

已知z=a+bi是虚数,且z+1/z是实数,求证:a^2+b^2=1
已知z=a+bi是虚数,且z+1/z是实数,求证:a^2+b^2=1

已知z=a+bi是虚数,且z+1/z是实数,求证:a^2+b^2=1
z+1/z =a+ib+1/(a+ib) =a+ib+(a-ib)/(a^2+b^2) =>[a+a/(a^2+b^2)]+i[b-b/(a^2+b^2)]是实数 =>[b-b/(a^2+b^2)]=0 =>a^2+b^2=1 ---------------- (z-1)/(z+1) =(a-1+ib)/(a+1+ib) =(a-1+ib)(a+1-ib)/(a+1+ib)(a+1-ib) =(a^2+b^2+2ib-1)/[(a+1)^2-b^2] 实部为=(a^2+b^2-1)/[(a+1)^2-b^2] =0/[(a+1)^2-b^2] =0 so （z-1)/(z+1)是纯虚数