If $f(z)g(z) = 0$ for every $z$, then $f(z) = 0$ or $g(z) = 0$ for every $z$.

Let's assume that $f$ is not identically $0$. Then there is a point $z_0\in\Omega$ such that $f(z_0)\neq0$. Since $f$ is continuous we find a neighbourhood of $z_0$ on which $f$ is free of zeros, $U$ say. But since $fg=0$ we have $g=0$ on $U$. Now use the identity principle to conclude $g=0$ in $\Omega$.