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Let $G$ be a region in $\mathbb{C}$ and $a\in G$. Suppose $f:G-\{a\}\to\mathbb{C}$ is an injective analytic function such that $f(G-\{a\})=\Omega$ is bounded. Show that $f(a)\in\partial\Omega$.

I know a couple things. Since $f$ is injective it's non-constant, so by the open mapping theorem we know $\Omega$ is open. Also, $f$ is obviously bounded in a neighborhood of $a$, so $z=a$ is a removable singularity. Then we can define $f$ so that it's analytic at $a$, possibly losing injectivity in the process. But I'm stuck here.

Any help is appreciated

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$f$ is open and injective, so you must show that the value $f(a)\notin \Omega$. –  Blah Mar 17 '12 at 18:41
    
Dear @Julián, I think the correct question is "Show that $f$ can be extended analytically through $a$ and that then $f(a)\in\partial\Omega$." –  Georges Elencwajg Mar 17 '12 at 20:24
    
@GeorgesElencwajg You must have written your comment in the few seconds it took me to delete it. –  Julián Aguirre Mar 17 '12 at 21:48

1 Answer 1

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As you have remarked the function $f$ can be extended to an analytical function $\tilde f:\ G\to{\mathbb C}$, and this extended function maps any neighborhood of $a$ onto a full neighborhood of $c:=\tilde f(a)$. It follows that for any $\epsilon>0$ the set $f\bigl(\dot U_\epsilon(a)\bigr)$ contains a punctured neighborhood of $c$.

Consider the points $z_n:=a+{1\over n}$ $\ (n\geq1)$. Since $f(z_n)\in\Omega$ for all $n\geq1$ and $c=\lim_{n\to\infty} f(z_n)$ it follows that $c\in\bar\Omega$.

Therefore we only have to exclude the case $c\in\Omega$. Assume to the contrary that there is a point $b\in G\setminus\{a\}$ with $f(b)=c$. Choose an $\epsilon>0$ such that $$\epsilon<{|b-a|\over 2},\quad U_\epsilon(a)\subset G, \quad U_\epsilon(b)\subset G\ .$$ Then $\dot U_\epsilon(a)$ and $U_\epsilon(b)$ are disjoint, and both their images under $f$ contain a full punctured neighborhood of $c$. It follows that $f$ would not be injective on $G\setminus\{a\}$.

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This is very nice. Thank you! –  Bey Mar 18 '12 at 21:36

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