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Complex translation length $\lambda$ is given by $tr X = 2\cosh \lambda,$ where $\Re \lambda > 0,$ and is the translation length, whilst the imaginary part is the rotation angle.

One proof is sketched here: http://www.maths.gla.ac.uk/wws/cabripages/hyperbolic/harea2.html A more brute force way of expressing area through side lengths is to use the hyperbolic law of cosines.

Yes. Cutting along the shortest geodesics $\alpha, \beta, \gamma$ between the pants boundaries produces two congruent right angled hexagons. They are congruent because a right angled hexagon is determ …

Yes, this is true for all of them. Any finitely generated matrix group has a torsion-free subgroup of finite index; this is the so-called "Selberg's lemma". A canonical source is Ratcliffe's Hyperboli …

There is a meaning, though whether it is geometric is up to you to decide. You can see it in: http://mathworld.wolfram.com/GeneralizedLawofSines.html which is fairly incomprehensible without http://ma …

The center of your group $G$ has to be finite (since it is finitely generated, abelian, and has no element of infinite order -- if it did, the group would fail to be word-hyperbolic, since that elemen …

The formula is: $S = \pi,$ thanks to Gauss-Bonnet.

Actually, pairs of pants (a.k.a. triply connected planar domains) are special, and it is unwise to derive results for them from "the general case". See this nice paper by T. Sugawa (1996) (for poster …

Yes, since the injectivity radius (defined as the max of injectivity radii over all points of the surface) is bounded by roughly the log of the area (think "embedded disk"). For interesting papers on …

G. Mostow and Y.-T. Siu, A compact Kahler surface of negative curvature not covered by the ball, Ann. Math. 112 (1980) 321-360

A very partial answer: by the results of Richard Bamler, bounds on Ricci curvature (which is less than what you are requiring here) and diameter (more than you are requiring), implies that the metric …

This question is studied by Ohshika and Potyagailo (here is a link to the lucid, as usual, mathreview by McCullough.) A more definitive result is due to T. Delzant and L. Potyagailo (which has nice re …

No, it is not. see, for example this reference: https://books.google.com/books?id=3s4bCAAAQBAJ&pg=PA101&lpg=PA101&dq=Weeks+manifold+haken&source=bl&ots=jCQ_jlddLA&sig=QMTywVy5BPwqsLnCPduk0XSzRuA&hl=en …

The magic words are "cone type inequality". For a quite nice discussion of this and related subject (with copious references) see Wenger's thesis.

You are absolutely right, your method does the trick.