z=[(i+1)^2+3(1-i)]/(2+i)
=(-1+2i+1+3-3i)/(2+i)
=(3-i)/(2+i)
=(3-i)(2-i)/5
=(5-5i)/5
=1-i
z^2+az+b=1+i
(1-i)²+a(1-i)+b=1+i
1-2i-1+a-ai+b=1+i
(-2-a)i+(a+b)=1+i
∴-2-a=1 a+b=1
解得a=-3,b=4