TITLE: Given two holomorphic functions on a region find two other such that...
QUESTION [7 upvotes]: Let $\Omega$ be a region in the complex plane and let $f_1$ and $f_2$ be holomorphic functions on $\Omega$ having no common zero. Show that there exist holomorphic functions $g_1$ and $g_2$ on $\Omega$ such that, $f_1g_1+f_2g_2$ is identically equal to one on $\Omega$.

REPLY [4 votes]: This is a special case of Theorem 15.15 in Rudin's Real and Complex Analysis, as Malik Younsi said. I'll simplify the proof for this case. 
Our goal is to find a holomorphic function $g_1$ such that $(1-f_1g_1)/f_2$  is holomorphic. Enumerate the distinct zeros of $f_2$ as $\{z_n:n=1,2,\dots\}$ with orders $\{m_n:n=1,2,\dots\}$. We need $1-f_1g_1$ to vanish at every $z_n$ to the order at least $m_n$. Formally, we need
$$1-f_1(z)g_1(z) = O((z-z_n)^{m_n+1}), \quad z\to z_n \tag{1}$$
Condition (1) is fulfilled by choosing $g_1$ such that 
$$
g_1(z) = \frac{1}{f_1(z)} + O((z-z_n)^{m_n+1}), \quad z\to z_n \tag{2} 
$$
for every $n$. The latter is made possible  by the  Mittag-Leffler interpolation theorem (note that $1/f_1$ is holomorphic at $z_n$).