The question
Let $f_1,f_2$ be some entire functions without zeros in common, so for every $z∈ℂ$ we have $|f_1(z)|^2+|f_2(z)|^2≠0$. Prove that there exist two entire functions $g_1,g_2$ such that:
$$ f_1(z)g_1(z) + f_2(z)g_2(z) = 1,\qquad ∀z ∈ \mathbb{D} $$
What I've tried
Lets define: $$ g_1 = \frac{1-g_2f_2}{f_1}$$
If we show that the fraction is entire, then $g_1$ is given as above. But because $f_1$ can be zero at some point, we need to choose $g_2$ correctly to have zeroes where $f_1$ does. In other words: $$ Z(f_1) ⊂ Z(1-g_2f_2) $$
Suppose $f_1(a)=0$, then we need to take $g_2(a)=1/f_2(a)$, and we have no problem with the denominator, since $Z(f_1)\cap Z(f_2)= \emptyset$. Using Mittag-Leffler's interpolation, one can find $g_2$.
My doubts [based on what 308745 says]
As said in https://math.stackexchange.com/a/876453/308745, the multiplicity of the zeroes at $f_1$ are a restriction that one needs to deal with. Finally one has to choose:
$$g_2(z) = \frac{1}{f_2(z)} + O((z-z_n)^{m_n+1}), \quad z\to z_n $$
where $z_n$ is a zero of $f_1$ with multiplicity $m_n+1$. But the Mittag-Leffler's corollary that I have, only states the following:
let $\{a_n\}_{n≥1}$ and $\lim_{n\to∞}|a_n|=∞$, and $\{w_n\}_{n≥1}$ be an arbitrary sequence of complex numbers. Then there exists an entire function $f$ such that $f(a_n)=w_n,\ ∀n≥1$
But in this problem I also need $(1-g_2f_2)^{(j)}(z_n)=0$ for $j≤m_n$.
How can I construct $(1-g_2f_2)$ to satisfy that?
