z=(1+i)平方+3(1-i)/2+i
=1^2+i^2+2i+3/2-3i/2+i
=3i/2+3/2
z^2+mz+n
=9(i^2+1+2i)/4+3m(1+i)/2+n
=9i/2+3m/2+3mi/2+n
=(9+3m)i/2+3m/2+n=1+i
所以,比较上面的到
9+3m=2
3m/2+n=1
解得,
m=-7/3
n=9/2