已知z=a+bi是虚数,且z+1/z是实数,求证：（z-1)/(z+1)是纯虚数

楼上强人
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