Editing Appendix/Ramblings/T1Coordinates (section)

===Coordinate Inversion===
Can we invert these expressions to obtain <math>\varpi</math> and <math>\Zeta</math> (or, equivalently, <math>z</math>) strictly in terms of <math>\chi_1</math> and <math>\chi_2</math>?  Well, we can start by squaring both coordinates and subtracting to isolate <math>\varpi^2</math>.  Specifically,
<div align="center">
<math>
\biggl( \frac{\chi_1}{B\varpi} \biggr)^2 - \biggl( \frac{\chi_2}{A\varpi^{1-q^2}} \biggr)^2 = \cosh^2\Zeta - \sinh^2\Zeta = 1 .
</math>
</div>
Hence, the desired functional expression <math>\varpi^2(\chi_1,\chi_2)</math> comes from the root(s) of the following polynomial equation:
<div align="center">
<math>
\biggl( \frac{\chi_2}{A} \biggr)^2 (\varpi^2)^{q^2} + \varpi^2 - \biggl(\frac{\chi_1}{B}\biggr)^2 = 0 .
</math>
</div>
It seems unlikely that closed-form analytic expressions can be extracted for the root(s) of this equation for an arbitrary choice of the flattening index, <math>q</math>.  But it should be possible to obtain analytically expressible roots in the case of <math>q^2 = 2</math>, <math>q^2 = 3</math>, and <math>q^2 = 4</math>.  Once the expression for <math>\varpi^2(\chi_1,\chi_2)</math> is known, a determination of <math>z(\chi_1,\chi_2)</math> is straightforward.  Specifically,
<div align="center">
<math>
z = \frac{\varpi}{q} \sinh\Zeta = \frac{\varpi}{q} \biggl[ \frac{\chi_2}{A} ~ \varpi^{q^2-1} \biggr] = \biggl[ \frac{\chi_2}{qA} ~ \varpi^{q^2} \biggr] .
</math>
</div>
In addition, from the above polynomial expression, we know that,
<div align="center">
<math>
\chi_2^2 (\varpi^2)^{q^2} = A^2 \biggl(\frac{\chi_1^2}{B^2} -  \varpi^2 \biggr) .
</math>
</div>
Hence, quite generally we deduce,
<div align="center">
<math>
z^2 = \frac{1}{q^2} \biggl(\frac{\chi_1^2}{B^2} -  \varpi^2 \biggr) ,
</math>
</div>
which, in retrospect, could have been derived more directly from the definition of <math>\chi_1</math>.