[(1+i)2(3+4i)2]/（2z）=2i*(24i-7)/2z=-48-14i/2z=-24-7i/z.(1)
设Z=a+bi
则：根号（a^2+b^2）=1+3i-a-bi,求出：b=3,a=-4,即：z=-4+3i,代入（1）式子中,得到：
=(24+7i)/(4-3i)=(24+7i)(4+3i)/25=3+4i