Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 4 down vote accepted
+50

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 $E_1$.

share|improve this answer
    
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
    
@RobertIsrael I believe that typically, one of the components is unbounded. Can your proof be extended in that case? –  user1337 Jul 30 '13 at 21:39
    
Use a fractional linear transformation to reduce to the case where they are bounded. –  Robert Israel Jul 31 '13 at 4:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.