已知复数z1满足(2+i)z1=3-i,z2=a+1+(a-2)i且|z1-{z2}|

设z1=a+bi ,(2+i)z1=3-i=（2a-b)+(a+2b)i
2a-b=3 a+2b=-1,a=1,b=-1,z1=1-i{z2}=a+1+（2-a)i
z1-{z2}=-a+(a-3)i |z1-{z2}|=√a^2+(a+3)^2
3+z1=4-i|3+z1|=√17
所以}√a^2+(a+3)^2