Surjectivity of a natural ring homomorphism

Hi,

I am stuck on a question, I think I know the process required to solve it as we covered a similar example in class. I'm just a bit stuck...

Let x=a+bi contained in the Gaussian integers, where a,b are coprime integers. I need to prove that the natural ring homomorphism

f:Z --> Z[i]/(x) given by f(n) = n + (x) is surjective.

So, if I have this right, I want to show that for all a,b in the natural numbers that a+ib+(x) =f(z) for some z contained in the natural numbers and I need to find i+(x) = f(z) = z+i

i.e. that z-i is contained in (x). But what can z be? My problem I think is that I still don't fully understand what (x) is. All I know is that

(x) = {(x)(a+ib) s.t a,b are integers} but I'm really confused!

Any help/advice would be great! Thanks,

Sooz