z=[(1+i)^3(a-i)^2]/[(√2(a-3i)^2]
所以
|z|=|(1+i)|^3*|(a-i)|^2/(√2|a-3i|^2)=2/3
(√2)^3*(a²+1)/[√2(a²+9)]=2/3
2(a²+1)/(a²+9)=2/3
a²+9=3a²+3
a=±√3
z=[(1+i)^3(a-i)^2]/[(√2(a-3i)^2]
所以
|z|=|(1+i)|^3*|(a-i)|^2/(√2|a-3i|^2)=2/3
(√2)^3*(a²+1)/[√2(a²+9)]=2/3
2(a²+1)/(a²+9)=2/3
a²+9=3a²+3
a=±√3