z=a+bi
在单位圆上
模等于1,所以a^2+b^2=1
(z^2+1)/z
=z^2/z+1/z
=z+1/z
=a+bi+1/(a+bi)
=a+bi+(a-bi)/(a+bi)(a-bi)
=a+bi+(a-bi)/(a^2+b^2)
=a+bi+(a-bi)/1
=2a
所以是实数,但不一定是0
所以选C
z=a+bi
在单位圆上
模等于1,所以a^2+b^2=1
(z^2+1)/z
=z^2/z+1/z
=z+1/z
=a+bi+1/(a+bi)
=a+bi+(a-bi)/(a+bi)(a-bi)
=a+bi+(a-bi)/(a^2+b^2)
=a+bi+(a-bi)/1
=2a
所以是实数,但不一定是0
所以选C