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I was recently reading about Laurent series for complex functions. I'm curious about a seemingly similar situation that came up in my reading.

Suppose $\Omega$ is a doubly connected region such that $\Omega^c$ (its complement) has two components $E_0$ and $E_1$. So if $f(z)$ is a complex, holomorphic function on $\Omega$, how can it be decomposed as $f=f_0(z)+f_1(z)$ where $f_0(z)$ is holomorphic outside $E_0$, and $f_1(z)$ is holomorphic outside $E_1$? Many thanks.

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    By hypothesis $\Omega$ is outside both $E_0$ and $E_1$ and so setting $f_{0,1}:=\frac{1}{2}f$ would work. Are you sure this is phrased correctly?2012-04-13
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    Dear @anon, I was paraphrasing to put it in question form, so I'll post verbatim what I was reading: Let $\Omega$ be a doubly connected region whose complement consists of the components $E_1$, $E_2$. Prove that every analytic function $f(z)$ in $\Omega$ can be written in the form $f_1(z)+f_2(z)$ where $f_1(z)$ is analytic outside of $E_1$ and $f_2(z)$ is analytic outside of $E_2$.2012-04-13
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    Maybe invoke analytic continuations into the two extended domains $\Omega\cap E_1$, $\Omega\cap E_0$ and average them?2012-04-13
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    @anon But aren't $\Omega\cap E_i=\emptyset$ in both cases? Unless you mean $\Omega\cup E_i$?2012-04-13
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    Sorry, yeah I meant $\cup$.2012-04-13
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    @anon So the idea is to extend $f$ analytically to some functions $g$ on $\Omega\cup E_0$ and $h$ on $\Omega\cup E_1$, and then take $f_0=\frac{g}{2}$ and $f_1=\frac{h}{2}$?2012-04-13
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    If $f$ can be extended like this then it looks like it would work. I'm not sure if this is the intention behind the exercise though, it's just the only thing I can think of.2012-04-13

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I'll suppose both $E_0$ and $E_1$ are bounded. Let $\Gamma_0$ and $\Gamma_1$ be disjoint positively-oriented simple closed contours in $\Omega$ enclosing $E_0$ and $E_1$ respectively, and $\Gamma_2$ a large positively-oriented circle enclosing both $\Gamma_0$ and $\Gamma_1$. Let $\Omega_1$ be the region inside $\Gamma_2$ but outside $\Gamma_0$ and $\Gamma_1$. Then for $z \in \Omega_1$ we have by Cauchy's integral formula, $$ f(z) = \frac{1}{2\pi i} \left( \int_{\Gamma_2} \frac{f(\zeta)\ d\zeta}{\zeta - z} - \int_{\Gamma_0} \frac{f(\zeta)\ d\zeta}{\zeta - z} - \int_{\Gamma_1} \frac{f(\zeta)\ d\zeta}{\zeta - z} \right)$$

If you're not familiar with this version of Cauchy's formula, you can draw thin "corridors" connecting $-\Gamma_0$, $-\Gamma_1$ and $\Gamma_2$ into a single closed contour enclosing $z$.

If $$f_k(z) = \frac{1}{2\pi i} \int_{\Gamma_k} \frac{f(\zeta)\ d\zeta}{\zeta - z}$$ this says $f(z) = f_2(z) - f_0(z) - f_1(z)$, where $f_2(z)$ is analytic everywhere inside $\Gamma_2$, $f_0(z)$ is analytic everywhere outside $\Gamma_0$, and $f_1(z)$ is analytic everywhere outside $\Gamma_1$. Moreover, the values of $f_k(z)$ don't depend on the choice of contours, as long as $z$ is inside $\Gamma_2$ and outside $\Gamma_0$ and $\Gamma_1$. By making $\Gamma_2$ sufficiently large and $\Gamma_0$ and $\Gamma_1$ sufficiently close to $E_0$ and $E_1$, any point in $\Omega$ can be included. So we actually have $f(z) = f_2(z) - f_0(z) - f_1(z)$ everywhere in $\Omega$, with $f_2(z)$ entire, $f_0(z)$ analytic outside $E_0$ and $f_1(z)$ analytic outside $E_1$.

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    Thanks Robert. So is it actually necessary to have that entire $f_2(z)$ function in the decomposition as well?2012-04-16
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    $f_2$ could be combined with either $f_0$ or $f_1$, according to taste.2012-04-16
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    @RobertIsrael I believe that typically, one of the components is unbounded. Can your proof be extended in that case?2013-07-30
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    Use a fractional linear transformation to reduce to the case where they are bounded.2013-07-31