# A more general case of the Laurent series expansion?

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? –  anon Apr 13 '12 at 5:31
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 $E_2$. –  Dedede Apr 13 '12 at 6:23
Maybe invoke analytic continuations into the two extended domains $\Omega\cap E_1$, $\Omega\cap E_0$ and average them? –  anon Apr 13 '12 at 7:57
@anon But aren't $\Omega\cap E_i=\emptyset$ in both cases? Unless you mean $\Omega\cup E_i$? –  Dedede Apr 13 '12 at 8:46
Sorry, yeah I meant $\cup$. –  anon Apr 13 '12 at 9:15

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 $E_1$.
Thanks Robert. So is it actually necessary to have that entire $f_2(z)$ function in the decomposition as well? –  Dedede Apr 16 '12 at 0:42
$f_2$ could be combined with either $f_0$ or $f_1$, according to taste. –  Robert Israel Apr 16 '12 at 1:23