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This is a classical theorem: hyperbolic Riemann surfaces of finite hyperbolic area are compact surfaces with finitely many punctures. Tsuji (Theorem XI.12) credits this to Siegel (1945). The proof is …

For statement 2, you have to specify whether $f$ is allowed to have zeros. (If yes, this is a metric with isolated singularities, but can be complete. If $f$ is free of zeros, it is always incomlete b …

Alexandr Solynin told me that he solved this problem (even the more general one, for hyperbolic n-gons with all zero angles) in 1993. A. Solynin, Some extremal problems for circular polygons, (Russian …

MR0725161 Ahlfors, Lars V. Möbius transformations in several dimensions, Minneapolis, MN, 1981.

Yes, this is a theorem of Siegel, and a reference with simple proof is Theorem XI.12 in M. Tsuji, Potential theory in modern function theory, Maruzen, Tokyo 1959 (there is an AMS Chelsea reprint, 1975 …

Here are some recent papers: Rafe Mazzeo, Hartmut Weiss arXiv:1509.07608 Teichmüller theory for conic surfaces arXiv:1710.09781 Rafe Mazzeo and Xuwen Zhu, Conical metrics on Riemann surfaces, I: th …

Your broken line closes iff the triangles $(v_1,v_2,v_3)$ and $(v_3,v_4,v_5)$ are equal, and $[v_1,v_3]$ is their common side. So there are two conditions: one is that $m_1=m_3$, by the hyperbolic law …

For the equivalence of definitions of quasiconformal maps the reference is J. Heinonen, Lectures on analysis on metric spaces, Springer 2001. Notice that the $K$ in the definiton you cite is not the s …

Yes, the answer is complicated, it is related to holomorphic dynamics, and the question was much studied, see for example: MR0869581 Lyubich, M. Yu.; Suvorov, V. V. Free subgroups of SL2(C) with two …

This paper is probably relevant for your question: MR0528966 Wolpert, Scott The length spectra as moduli for compact Riemann surfaces. Ann. of Math. (2) 109 (1979), no. 2, 323–351. And here is a more …

A simply connected region with 4 marked boundary points is called a quadrilateral. A conformal map of quadrilaterals is a conformal map of the regions which sends the marked boundary points to the mar …

Let $\{ z_1,\ldots,z_n\}$ be any finite set, and $$L_k(z)=\prod_{j\neq k}(z-z_j).$$ Then $L_k,\; 1\leq k\leq n$ are linearly independent since the set of their linear combinations consists of all poly …

W. Thurston, Three-dimensional geometry and topology.

In your statement you said "let S be a surface". First of all one has to be careful about what is a "surface". It must be a) orientable, and b) separable (have a countable base of topology). (All Rei …

Let me add to this nice answer of Ian Agol, that the isometry group is always finite, and contains at most $84(g-1)$ elements, if we count orientation-preserving (conformal) isometries, where $g\geq 2 …