Editing Appendix/Ramblings/T1Coordinates (section)

===First Special Case (quadratic)===
When <math>q^2 = 2</math>, the above polynomial becomes a quadratic equation and therefore its roots can be determined analytically.  Setting,
<div align="center">
<math>
x \equiv \varpi^2 ,
</math>
</div>
the relevant quadratic equation is,
<div align="center">
<math>
\biggl( \frac{\chi_2}{A} \biggr)^2 x^2 +  x - \biggl( \frac{\chi_1}{B} \biggr)^2 = 0 .
</math>
</div>
The root of this function that provides positive values for <math>x</math> (''i.e.,'' for <math>\varpi^2</math>) is,
<div align="center">
<math>
x = \frac{A^2}{2\chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] ,
</math>
</div>
where,
<div align="center">
<math>
\Chi_\mathrm{quad} \equiv \frac{2\chi_1 \chi_2}{AB} = \sinh (2\Zeta).
</math>
</div>
This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 2</math>.  The desired expression for <math>z(\chi_1,\chi_2)</math> is,
<div align="center">
<math>
z^2 =  \frac{\chi_1^2}{q^2 B^2}- \frac{A^2}{2q^2 \chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] = \frac{A^2}{4 \chi_2^2}\biggl[ \frac{1}{2} \Chi_\mathrm{quad}^2 +1 -  \sqrt{1 + \Chi_\mathrm{quad}^2}\biggr] .
</math>
</div>

We note as well that in this special case, where <math>q^2 = 2</math>, 
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
4 L^2 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\varpi^2 \biggl[1 + 2 \sinh^2 \Zeta  \biggr] = \varpi^2 \cosh(2\Zeta) = \varpi^2 \biggl[ 1+\sinh^2(2\Zeta) \biggr]^{1/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{A^2}{2\chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] \sqrt{ 1 + \Chi_\mathrm{quad}^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{A^2}{2\chi_2^2} \biggl[ \Chi_\mathrm{quad}^2 + 1 - \sqrt{ 1 + \Chi_\mathrm{quad}^2} \biggr] .
</math>
  </td>
</tr>
</table>
Note that this expression is very similar, but not identical, to the expression just derived for <math>z^2</math>.