# $U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$?

Let $U \subseteq \mathbb{C}$ be open. I want to construct a holomorphic function $f: U \to \mathbb{C}$, such that for all $z \in \partial U$ and for all $\varepsilon > 0$, there is no holomorphic continuation $\tilde{f}: U \cup B_\varepsilon(z) \to \mathbb{C}$ of $f$. For simply connected $U$, I could just find a diffeomorphism to the unit circle and there, the geometric series would be an example of such a function, if I am not mistaken. But what am I supposed to do if $U$ isn't simply connected? The Laurent-series comes to mind and also is the current topic of our complex-analysis lecture, however what would I do if those holes of $U$ were not "circle-shaped"?

Thanks for any help in advance.

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The geometric series has an analytic continuation at all points on the unit circle except $z = 1$, given by $\tilde{f}(z) = \frac1{1-z}$. – Lukas Geyer Dec 6 '12 at 20:43
@LukasGeyer: I was not aware of that. I thought since the geometric series diverges on the boundary of its convergence radius, it can't be analytically continued. Then now I am even more helpless than I was before. – Huy Dec 6 '12 at 20:50
As a hint: You can find solutions of the form $\sum_k \frac{a_k}{z-z_k}$, where $(z_k)$ is dense in the boundary of $U$, and $f(z) \to \infty$ as $z\to z_k$. – Lukas Geyer Dec 6 '12 at 20:56