设z1=a+2i ,z2=3-4i,z1/z2=(a+2i)/(3-4i)=(a+2i)(3+4i)/25=(3a+6i+4ai-8)/25=(3a-8)/25+(4a+6)i/25,(3a-8)/25=0,a=8/3,4*8/3+6=50/3≠0，∴z1=8/3+2i.
8/3+2i
