z=[(1+2i-1)+(3-3i)]/(2+i)
=(3-i)/(2+i)
=(3-i)(2-i)/[(2+i)(2-i)]
=(6-3i-2i-1)/(4+1)
=(5-5i)/5
=1-i
z²+ax+b=1+i
1-2i-1+a-ai+b=1+i
(a+b)+(-2-a)i=1+i
所以a+b=1
-2-a=1
a=-3
b=4