I have a polynomial $f\in\mathbb{C}[X\_1,...X\_n]$ and a bounded (non-empty) compact region $\Omega \subset \mathbb{C}^n$. Let's say additionally that $f$ is not zero on the boundary of $\Omega$.
Does there exist a certificate/criterion, which tells me whether there exists an $a\in\Omega$ such that $f(a)=0$?


I think for the univariate case, one could apply the [Argument Principle](https://en.wikipedia.org/wiki/Argument_principle).



Certificate/Criterion for the existence of zero of a complex multivariate polynomial in a bounded region
So, I think Zach is right, though you need a little more care. Since $\partial\Omega$ is not algebraic, it is not entirely straightforward that any hypersurface should intersect it, so one would need a version of the Jordan–Brouwer separation theorem to conclude that if there is a zero inside, then there has to be one on the boundary as well. This would boil down to analyzing the shape of $\Omega$, which could probably be made nice using the fact that $f$ is a polynomial. On the other hand, here is an alternative argument which shows that $f$ doesn't even have to be a polynomial. If it is holomorphic everywhere in $\Omega$, that's enough.


First, the multivariate case:


Just to make sure, let us assume that $\partial\Omega$ is connected. It seems to me that one can reduce to this case (especially if we are dealing with a polynomial), but sometimes these "simple" topological issues take a long time to sort out, so let me leave this to the reader. 



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> > **Claim.** Let $f$ be a holomorphic function on $\Omega$ (whose boundary is connected). 
> >  If $n\geq 2$ and $\forall z\in\partial\Omega$, $f(z)\neq 0$, then $\forall z\in\Omega$, $f(z)\neq 0$.
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**Proof:** Let $g=\dfrac 1f$. Then $g$ is a holomorphic function in a neighbourhood of $\partial\Omega$, so by Hartogs' theorem it admits a holomorphic extension to $\Omega$ which we will also denote by $g$. So, $g$ and $f$ are two holomorphic functions on $\Omega$ and $fg=1$, so neither can have a zero in $\Omega$. $\square$ 


OK, so this leaves the $n=1$ case, which is indeed settled by the Argument Principle as you suggest. (Since $f$ is holomorphic inside $\Omega$, it cannot have a pole, so the integral that appears in the Argument Principle would be non-zero exactly if $f$ had a zero.)


