First of all, note that Q(i) is not a complex number field, but a subdomain, and there may be many automorphisms of complex number fields (depending on the axiom of selection).

If you want to define the number of automorphisms, the most basic thing is to know that identity maps are automorphisms, and then you only need to find non-identity maps.

Firstly, it is proved that the automorphism of Q has only one unit mapping, because f(0)=0 and f( 1)= 1. Then, for positive integers m, n,

F(mx)=f(x)+...+f(x)=mf(x), and it can be deduced that f(m)=mf( 1)=m,

Then f (1) = f (n/n) = nf (1/n) gives f( 1/n)= 1/n, so f(m/n)=m/n,

Finally, x> is at 0 and f (-x) = f (0)-f (x) =-x.

If f is an automorphism of Q(i), then according to the above discussion, f must be an automorphism on q. In this case,-1=f(- 1)=f(i)f(i), it can be obtained that f(i)=i or f(i)=-i, and f is composed of f (.

Don't say another question, you can do it yourself, even asking such a question is mindless.