设z=a+bi
/z/（z的模）=1+3i-z =(1+a)+(3-b)i
因为|z|是实数,不是复数
3-b=0 b=3
|z|=√(a^2+b^2)=(1+a)
a=4
z=4+3i
1/z=(4-3i)/25
(1+3i)(1+3i)(3+4i)(3+4i)
=(-8+6i)(-7+24i)
=88-234i
(1+3i)(1+3i)(3+4i)(3+4i)/2z
=(88-234i)(4-3i)/50
=-7-20i