The problem: Given entire functions $f,g : \mathbb{C} \to \mathbb{C}$, such that all the zeroes of both functions are contained in a disk $D_r(0)$ and they have an equal number of them (counting multiplicities). Prove that there exists an entire and non-vanishing function $h : \mathbb{C} \to \mathbb{C} \setminus \{0\}$ and a radius $R > r$ such that $$|f(z) - g(z) h(z)| < |f(z)|$$ on the circle $|z| = R$.
What I've tried:
Applying the argument principle to both $f$ and $g$, and subtracting both integrals, I was able to show that $\int_C \frac{(f/g)'}{f/g} dz = 0$ for all circles $C$ sufficiently large (in $\Omega = \mathbb{C} \setminus (\text{cl }D_r(0)).$ From this it follows that I can take the logarithm, i.e., find a holomorphic function $H : \Omega \to \mathbb{C}$ such that $f/g = e^H$ on $\Omega$. But then I got stuck. It seems that I might be able to expand $e^H$ as a power series which converges uniformly on $|z| = R$ hence we might pick a large enough polynomial in that series and call it $h$. But the problem is that I don't see why these terms in the series have to be holomorphic everywhere.
Some help would be appreciated! Full solution is allowed, honestly I'm kind of sick of trying to solve this one...