Inverse Rouché problem?

The problem on one of my mock-exams is as follows: Suppose both $f$ and $g$ are entire with both a finite amount of zeros that all lie in a disk $D_r(0)$. The zeros are counted by their multiplicity. Suppose that the amount of zeros of both function is is equal. Prove that there exists an entire function $h$ that has no zeros such that there is a sufficiently large $R > r$ for which it holds that $|f(z) - h(z)g(z)| < |f(z)|$ for all $|z|=R$.

A sub question I have to this is why do we need the condition that the amount of zeros is equal? Why can $h$ and $R>r$ not exist if the number of zeros aren't equal?

I suppose this has to to with some kind of inverse of Rouché's theorem, because we're starting from to fact that the number of zeros is equal and we want to prove an inequality.

$\textbf{Rouché's Theorem:}$ Suppose $f$ and $g$ are holomorphic in an open set containing a circle $C$ and its interior. If $|f(z)| > |g(z)|$ for all $z \in C$ then $f$ and $f+g$ have the same amount of zeros inside the circle $C$.

Could anyone give a hint? Thanks!

This problem is similar to Inverse statement to Rouché's theorem in complex analysis.

Solution 1:

Assume first $f=P, g=Q$ non-zero polynomials of the same degree $n \ge 0$ (by hypothesis); if $a \ne 0$ is the ratio of their leading coefficients, $f-ag$ has degree at most $n-1$ (which for example means $f-ag=0$ if $n=0$), so $|f-ag|<|f|$ on a large enough circle $|z|=R$ and we are done.

But now the general case easily follows as $f,g$ entire with finitely many zeroes, means that $f=Pe^{f_1}, g=Qe^{g_1}$ with $P,Q$ non zero monic polynomials of the same degree and $f_1,g_1$ entire (just factor out the finitely many zeroes from $f,g$ with the corresponding monic polynomials and what remains are never vanishing entire functions, so are exponentials of entire functions), so taking $h=e^{f_1-g_1}$ one has $|f-hg|=|e^{f_1}(P-Q)|<|e^{f_1}P|$ for a large enough circle since $P-Q$ has degree strictly less than $P$