# Biholomorphic functions and delaunay triangulation

Lets have a look at the two simply connected domains $$D,G \subset \mathbb{C}$$ and a biholomorphic function $$f:D \rightarrow G$$ which maps $$D$$ conformal onto $$G$$.

For some $$n \in \mathbb{N}$$ there are points $$b_k \in B_n := \lbrace b_d \in D \mid d = 1,...,n \rbrace$$ such that the delaunay triangulation $$\Delta_n(B_n)$$ approximates the domain $$D$$.

I know that the delaunay triangulation is not unique but lets map all those points $$B_n$$ onto $$G$$:

Let $$f(B_n)$$ bei the set of the correspondenting points in $$G$$. Lets connect all the points in $$f(B_n)$$ the same way we did in $$\Delta_n(B_n)$$. We get a triangulation $$\Delta_n(f(B_n)) for$$G$. I ask myself if this triangulation is a valid delaunay triangulation again. In other words: keeps the validation of a delaunay triangulation stable under conformal mappings? ## 1 Answer It's very unlikely that a Delaunay triangulation is stable under a conformal mapping. Try for instance the map $$z \mapsto 1/z$$ on a disc not containing the origin. Let $$B=4$$ be the vertices of a triangle and its barycenter. One chance of preserving Delaunay triangulation would be if the mapping preserves circles. Perhaps it would be stable under a Möbius transformation, though I doubt it. • But$z \mapsto z^2\$ is not a conformal mapping. – Arjihad Nov 10 '18 at 18:34
• Yes. I should have mentioned that. Im considering conformal mappings to be biholomorphic funtions. – Arjihad Nov 10 '18 at 18:55