(a+i)/(2-i)
=[(a+i)(2+i)]/[(2+i)(2-i)]
=(2a+ai+2i+i²)/(4-i²)
=[(2a-1)+(a+2)i]/5
=(2a-1)/5+(a+2)i/5
∵是纯虚数
则实部为0
∴(2a-1)/5=0
即a=1/2