Editing Appendix/Ramblings/PPToriPt1A (section)

====Our Notation====
As is explicitly defined in [[Appendix/Ramblings/AzimuthalDistortions#Figure1|Figure 1 of our accompanying detailed notes]], we have chosen to represent the spatial structure of an eigenfunction in the equatorial-plane of toroidal-like configurations via the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ f_m(\varpi)e^{-im\phi_m} \biggr\}  \, .</math>
  </td>
</tr>
</table>
</div>
In general, we should assume that the function that delineates the radial dependence of the eigenfunction has both a real and an imaginary component, that is, we should assume that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_m(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}(\varpi) + i\mathcal{B}(\varpi) \, ,</math>
  </td>
</tr>
</table>
</div>
in which case the square of the modulus of the function is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~|f_m|^2 \equiv f_m \cdot f^*_m </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}^2 + \mathcal{B}^2 \, .</math>
  </td>
</tr>
</table>
</div>
We can rewrite this complex function in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_m(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|f_m|e^{-i[\alpha(\varpi) + \pi/2]} \, ,</math>
  </td>
</tr>
</table>
</div>
if the angle, <math>~\alpha(\varpi)</math> is defined such that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sin\alpha = \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\cos\alpha = \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl(\frac{\mathcal{A}}{\mathcal{B}}\biggr) 
= \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] 
\, .</math>
  </td>
</tr>
</table>
</div>

Hence, the spatial structure of the eigenfunction can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|f_m(\varpi)|e^{-i[\alpha(\varpi) + \pi/2+ m\phi_m]} \, . </math>
  </td>
</tr>
</table>
</div>

From this representation we can see that, at each radial location, <math>~\varpi</math>, the phase angle(s) at which the fractional perturbation exhibits its maximum amplitude, <math>~|f_m|</math>, is identified by setting the exponent of the exponential to zero.  That is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\phi_m = \phi_\mathrm{max}(\varpi)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-\frac{1}{m}\biggl[\alpha(\varpi) +\frac{\pi}{2}\biggr] 
= -\frac{1}{m}\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} \biggr\}
\, .</math>
  </td>
</tr>
</table>
</div>

An equatorial-plane plot of <math>~\phi_\mathrm{max}(\varpi)</math> should produce the "constant phase locus" referenced, for example, in recent papers from the [[Appendix/Ramblings/ToHadleyAndImamura#Summary_for_Hadley_.26_Imamura|Imamura &amp; Hadley collaboration]].

<!-- SECOND ATTEMPT
This is the form that has been adopted broadly by the numerical simulation community, as graphical displays of <math>~f_m(\varpi)</math> and <math>~\phi_m(\varpi)</math> have been used to study the structure of unstable eigenmodes &#8212; see, for example, our discussion of [[Appendix/Ramblings/ToHadleyAndImamura#Summary_for_Hadley_.26_Imamura|simulation results published by the Imamura &amp; Hadley collaboration]].  Multiplying this expression through by its complex conjugate gives the square of the modulus of the function.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl| \frac{\delta\rho}{\rho_0}\biggr|^2_\mathrm{spatial} \equiv
\biggl[ \frac{\delta\rho}{\rho_0}\biggr] \cdot \biggl[ \frac{\delta\rho}{\rho_0}\biggr]^* 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~f^2_m(\varpi)e^{-im[\phi_m(\varpi)]} \cdot e^{+im[\phi_m(\varpi)]} = f^2_m(\varpi) \, .</math>
  </td>
</tr>
</table>
</div>
We see, therefore, that written in this manner, <math>~f_m</math> is the modulus of the eigenfunction.

Alternatively, we could choose to omit explicit reference to an angular phase function and write the perturbation amplitude as a function with an imaginary as well as a real part, say,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}(\varpi) - i\mathcal{B}(\varpi)  \, .</math>
  </td>
</tr>
</table>
</div>
This is the form often used in research papers that seek to find analytic expressions for the structure of unstable eigenmodes, such as the works of [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes85] and [http://adsabs.harvard.edu/abs/1986MNRAS.221..339G GGN86].  A mapping from one representation to the other is accomplished by, first, constructing the modulus of the complex perturbation amplitude and equating it to <math>~f_m</math>, that is,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_m(\varpi) = \sqrt{\biggl[ \frac{\delta\rho}{\rho_0}\biggr] \cdot \biggl[ \frac{\delta\rho}{\rho_0}\biggr]^* }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{\mathcal{A}^2(\varpi) + \mathcal{B}^2(\varpi)} \, .</math>
  </td>
</tr>
</table>
</div>
Second, the phase function is obtained via the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~m\phi_m(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[ \frac{\mathcal{B}(\varpi)}{\mathcal{A}(\varpi)} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
where, more thoroughly it must be understood that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cos(m\phi_m) = \biggl[ \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr]</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\sin(m\phi_m) = \biggl[ \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>


To demonstrate that this is the proper mapping for the phase function, let's plug the expression for <math>~f_m</math> along with the expressions for <math>~\cos(m\phi_m)</math> and <math>~\sin(m\phi_m)</math> into our first relation, that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~f_m(\varpi) \{ \cos[m\phi_m(\varpi)] - i\sin[m\phi_m(\varpi)] \}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{\mathcal{A}^2 + \mathcal{B}^2}\biggl\{ \biggl[ \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] 
- i\biggl[ \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \mathcal{A} - i\mathcal{B} \, .</math>
  </td>
</tr>
</table>
</div>

Q.E.D.
-->
<!-- USEFUL (BUT NOT FULLY CORRECT) MANIPULATION OF COMPLEX EIGENFUNCTION EXPRESSIONS ...

Recognizing that the leading factor, <math>~f_m</math>, is, in general, composed of both a real part and an imaginary part, we can rewrite this expression as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\mathrm{Re}(f_m) + i\mathrm{Im}(f_m)\biggr] \biggl[\cos(m\phi_m) - i\sin(m\phi_m) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)\biggr] 
+ i \biggl[- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} \biggl\{
\biggl[\frac{\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} }\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ i \biggl[\frac{- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} }\biggr] 
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \biggl\{\sin(\beta_f + m\phi_m) + i \cos(\beta_f + m\phi_m)\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>

where, the modulus of this function is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \equiv
\sqrt{\biggl[ \frac{\delta\rho}{\rho_0}\biggr]\cdot\biggl[ \frac{\delta\rho}{\rho_0}\biggr]^*}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{
\biggl[\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)\biggr]^2 
+ \biggl[- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)\biggr]^2
\biggr\}^{1/2}
</math> 
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{
[\mathrm{Re}(f_m) \cos(m\phi_m)]^2 + 2\mathrm{Re}(f_m)\mathrm{Im}(f_m)\sin(m\phi_m)\cos(m\phi_m)+ [\mathrm{Im}(f_m)\sin(m\phi_m)]^2
</math> 
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
+ [\mathrm{Re}(f_m)\sin(m\phi_m)]^2 - 2\mathrm{Re}(f_m)\mathrm{Im}(f_m)\sin(m\phi_m)\cos(m\phi_m) + [\mathrm{Im}(f_m)\cos(m\phi_m)]^2 \biggr\}^{1/2}
</math> 
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \sqrt{
[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} \, ,
</math> 
  </td>
</tr>
</table>
</div>

and where we have introduced a new angle, <math>~\beta_f</math>, such that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sin\beta_f = \frac{\mathrm{Re}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\cos\beta_f = \frac{\mathrm{Im}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \beta_f </math>
  </td>
  <td align="center">
<math>~\equiv</math> 
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[\frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Alternatively, we could choose to write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \biggl\{ \cos(\alpha_f - m\phi_m) + i \sin(\alpha_f - m\phi_m)\biggr\} 
= \biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} e^{i(\alpha_f -m\phi_m)} \, ,
</math>
  </td>
</tr>
</table>
</div>
where we have introduced a new angle, <math>~\alpha_f</math>, such that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cos\alpha_f = \frac{\mathrm{Re}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\sin\alpha_f = \frac{\mathrm{Im}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \alpha_f </math>
  </td>
  <td align="center">
<math>~\equiv</math> 
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[\frac{\mathrm{Im}(f_m)}{\mathrm{Re}(f_m)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Notice that, when written in this form, it is clear from taking the complex conjugate of the function that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha_f - m\phi_m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ m\phi_m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[\frac{\mathrm{Im}(f_m)}{\mathrm{Re}(f_m)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

FINISHED EXTRACTION OF COMPLEX FUNCTION MANIPULATION   -->