Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
$f$ 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
up vote 3 down vote accepted

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 $G\setminus\{a\}$.

share|cite|improve this answer
This is very nice. Thank you! – Bey Mar 18 '12 at 21:36

Your Answer


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.