若复数a+3i/1+2i(a属于R,i为虚数单位)是纯虚数,则a的值为

(a+3i)/(1+2i)
=(a+3i)(1-2i)/[(1+2i)(1-2i)]
=(a-2ai+3i+6)/(1+4)
=[a+6+(3-2a)i]/5
因为复数a+3i/1+2i(a属于R,i为虚数单位)是纯虚数,
所以a+6=0
3-2a≠0
解得a=-6
答案：a=-6非常感谢！不客气选为满意回答吧