Let \(f\) be holomorphic in \(0 < \abs{z-z_0} < r\), minus a sequence of poles \(\ts{z_k} \to z_0\). Show that for any \(w\in \CC\), there is a sequence \(\ts{w_k}\to z_0\) with \(f(w_k)\to w\).
Solution
Toward a contradiction, fix \(w\) and suppose no such sequence exists. Then for every sequence \(w_k\to z_0\), there is an \(\eps\) such that \(\abs{f(w_k) - w} \geq \eps\) for all \(k\). So the function \(g(z) \da {1\over f(z) - w}\) has no poles in \(D_r(z_0)\smts{z_0}\), and since each \(z_k\) is a pole of \(f\), each is a zero of \(g\). If \(z_0\) is a singularity, since \(\abs{g(z)} \leq \eps\), it is removable and thus \(g\) can be extended holomorphically over \(z_0\). By continuity, since \(z_k\to z_0\) with \(g(z_k) = 0\), we have \(g(z_0) = 0\). By the identity principle, \(g\equiv 0\), which means that every \(z\in D_r(z_0)\smts{z_0}\) is a zero of \(g\) and thus a pole of \(f\). But this contradicts that \(f\) is holomorphic on \(D_r(z_0)\smts{z_k}\). \(\contradiction\)