Let \(f\) be analytic in \(\Omega: 0<|z-a|<r\) except at a sequence of poles \(a_n \in \Omega\) with \(\lim_{n \rightarrow \infty} a_n = a\). Show that for any \(w \in \mathbb C\), there exists a sequence \(z_n \in \Omega\) such that \(\lim_{n \rightarrow \infty} f(z_n) = w\).
Explain the similarity and difference between the above assertion and the Weierstrass-Casorati theorem.
Solution
We prove the stronger conclusion that the sequence can be chosen with \(z_n\to a\).
Fix \(w\in\CC\). Suppose this stronger conclusion fails. Then there are \(\varepsilon>0\) and \(\delta>0\) such that \[ |f(z)-w|\ge\varepsilon \] whenever \(0<|z-a|<\delta\) and \(z\) is not one of the poles of \(f\). Indeed, otherwise for every \(n\) one could choose \(z_n\) with \(0<|z_n-a|<1/n\) and \(|f(z_n)-w|<1/n\).
On the punctured disk, away from the poles, define \[ G(z)={1\over f(z)-w}. \] Then \(|G(z)|\le\varepsilon^{-1}\). At each pole \(a_k\) of \(f\) lying in this disk, \(G(z)\to0\) as \(z\to a_k\), so \(a_k\) is a removable singularity of \(G\) and the extension satisfies \(G(a_k)=0\). After filling in all such points, \(G\) is holomorphic and bounded on \(0<|z-a|<\delta\). Hence \(a\) is also a removable singularity of \(G\).
Because \(a_k\to a\) and \(G(a_k)=0\), continuity of the extension gives \(G(a)=0\). Thus the zeros \(a_k\) of the extended holomorphic function \(G\) accumulate at the interior point \(a\). The identity theorem gives \(G\equiv0\) on the disk. This is impossible, since at every ordinary point where \(f\) is finite one has \(G=1/(f-w)\ne0\).
Therefore for every \(w\in\CC\) there exists a sequence \(z_n\to a\) with \(f(z_n)\to w\).
This resembles Casorati–Weierstrass because every finite complex value is a limit value arbitrarily near \(a\). The difference is that Casorati–Weierstrass concerns an isolated essential singularity, whereas here \(a\) is not an isolated singularity at all: poles of \(f\) accumulate at \(a\).