设 z=a+bi
∴√(a²+b²)=1+3i-a-bi=(1-a)+(3-b)i
∴a²+b²=(1-a)²
3-b=0
∴a=-4
b=3
∴2z/(1+i)²(3+4i)²
=2(-4+3i)/2i(24i-7)
=(-4+3i)/(-24-7i)
=(4-3i)(24-7i)/625
=(96-21-100i)/625
=(3-4i)/25