设z=x+iy, x,y为实数.由题意 lzl=1+3i-z,等式左边模长为实数,故等式右边虚数部分必抵消,否则等式不成立,可判断,y=-3,即z=x-3i,代回原等式,x^2+9=(x+1)^2,解得x=4故z=4-3i(1+i)^2(3+4i)^2/2z=2i*(3+4i)^2/[2*(4-3i)]=i*(3+4i)^2/(4-3i)=i*(3+4i)(3+4i)(4+3i)/25=(3+4i)(3+4i)(4i-3)/25=(3+4i)*(-25)/25=-3-4i
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