由(1+2i)z=4+ai,得:z= 4+ai 1+2i = (4+ai)(1?2i) (1+2i)(1?2i) = 4+2a+(a?8)i 5 = 4+2a 5 + a?8 5 i,∵复数z的实部与虚部相等,∴4+2a=a-8,解得:a=-12.故选:D.