10
$\begingroup$

The following is exercise 15 in section V.1 of Conway's Functions of One Complex Variable ("Classification of Singularities"). I'm currently studying for a complex analysis qualifying exam and this has appeared in the past.

Let $f$ be analytic in $G=\{z:0<|z-a|<r\}$ except that there is a sequence of poles $\{a_n\}$ in $G$ with $a_n\rightarrow a$. Show that for any $w$ in $\mathbb{C}$ there is a sequence $\{z_n\}$ in $G$ with $a=\lim z_n$ and $w=\lim f(z)$.

The conclusion makes me want to apply the Casorati-Weirstrass theorem. However, the singularity at $a$ is not isolated. As far as I know, an essential singularity is a particular type of isolated singularity. Am I wrong about this?

Any help would be greatly appreciated.

$\endgroup$
2
  • $\begingroup$ You are correct, an essential singularity must be isolated, so you can't apply Casorati-Weierstrass. $\endgroup$ – Robert Israel Apr 22 '12 at 19:30
  • 1
    $\begingroup$ Note that different authors use "essential singularity" with at least two non-equivalent definitions. Some require that an "essential singularity" must be isolated; others seem to use the term about any obstacle to analytic continuation that is not a pole. $\endgroup$ – hmakholm left over Monica Apr 22 '12 at 19:35
9
$\begingroup$

Hint: Suppose no such sequence exists.
Consider $g(z) = 1/(f(z) - w)$ (with $g(a_n) = 0$). What can you say about its singularity at $a$?

$\endgroup$
1
  • $\begingroup$ Thanks a lot for your help. I think I got it. $\endgroup$ – John Adamski Apr 22 '12 at 20:44
13
$\begingroup$

Define $V_\delta:=\{z\in G:|z-a|<\delta\}$. Fix $w$ in $\mathbb{C}$ and suppose that there is no such sequence. Then there exists an $\epsilon>0$ and $\delta>0$ such that $|f(z)-w|>\epsilon$ for all $z$ in $V_\delta\setminus\{a_n\}$. Now we can define the function $$g(z)=\frac{1}{f(z)-w},\quad z\in V_\delta$$ which is holomorphic throughout $V_\delta$ with zeros at each $a_n$.

Since $g$ is bounded by $1/\epsilon$, it follows that $g$ has a removable singularity at $a$. But since $a$ is a limit point for the zeros of $g$, the analytic continuation of $g$ to $V_\delta\cup\{a\}$ must have $g(a)=0$. Thus the set of zeros of the analytic continuation of $g$ has a limit point and therefore this analytic continuation must be identically zero. This contradicts the fact that $f$ is holomorphic in $G$ except at a sequence of poles. Therefore there must be such a sequence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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