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

解： (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