What we want to show is given u+iv we can find n so that n + (x) = u+iv + (x). This is equivalent to saying n - (u+iv) in (x) thus there exists s,t so that n - (u+iv) = (s+it)(a+bi) where a,b coprime (positive) integers. Thus, (n-u)+iv= (as - bt) + i(at+bs). Thus we want v = at+bs and n-u = as-bt. Since gcd(a,b)=1 it means there exists s,t so that v = at+bs. But then let n = u + (as - bt), which will make f(n) = u+iv + (x).

In general this is what we do: Given z+(x) in Z[i]/(x) let s,t be integrers such that at+bs = Im(z) then let n = Re(z)+(as-bt).