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Here's a (slightly paraphrased) excerpt from Complex Analysis by Stein & Shakarchi. The function $F:\mathbb D\to P$ is conformal where $P$ is a polygonal region (open, bounded, simply connected, polygonal boundary).

Lemma 4.4: Let $z_0$ be a point on the unit circle. Then $F(z)$ tends to a limit as $z$ approaches $z_0$ within the unit disc.

Proof. If not, there are two sequences $\{z_n\},\{z_n'\}$ in the unit disc that converge to $z_0$ and are so that $\{F(z_n)\},\{F(z_n')\}$ converge to two distinct points $\zeta,\zeta'\in\bar P$. ... We may therefore choose two disjoint discs $D,D'$ centered at $\zeta,\zeta'$, respectively. For all large $n$, $F(z_n)\in D, F(z_n')\in D'$. Therefore, there exist two continuous curves $\Lambda$ and $\Lambda'$ in $D\cap P$ and $D'\cap P$, respectively, with $F(z_n)\in\Lambda,F(z_n')\in\Lambda'$ for all large $n$, and with the end-points of $\Lambda$ and $\Lambda'$ equal to $\zeta,\zeta'$, respectively.

I wonder why the curves in the highlighted sentence would exist. I tried (by path-connectedness) to connect the $z_n$'s one by one, but the curve may not even have finite length (say if $|z_n-z_{n+1}|=1/n$). Does this use the assumption of $P$ being polygonal in an essential way? Or would simply connected with a "nice" boundary suffice? I think we are just working inside the disc $D$, but I'm not sure--maybe if $P$ has a weird shaped boundary, its intersection with $D$ would be too small to guarantee such curves.

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