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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 $\Omega$.

Dimitris
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    An obvious first idea is to consider $g_1=\frac{\overbar{f_1}}{|f_1|^2+|f_2|^2}$ and $g_2$ similarly. But are these functions holomorphic? – Dimitris Jul 10 '14 at 10:39
  • Your function is not going to be holomorphic because $g_1 = \dfrac{1}{f_1+\frac{f_2 {f_2}^}{{f_1}^}}$ (ignoring the poles). – user21820 Jul 10 '14 at 11:50
  • I don't have the book with me right now but a proof of this can be found in Rudin's real and complex analysis, chapter 15 if I recall correctly. – Malik Younsi Jul 12 '14 at 12:01

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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 $z_n$).