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Notebook[{

Cell[CellGroupData[{
Cell["Lange's Converse to Stern and Stolz", "Title"],

Cell["\<\
Jerry Lange's improved version of Van Vleck's converse to the theorem of \
Stern and Stolz\
\>", "Subtitle"],

Cell[TextData[{
  "The proof of Theorem 12.3 on page 223 includes the assertion that if \
A,A',B,B' are complex numbers such that ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{"|", 
            RowBox[{"det", "(", GridBox[{
                  {"A", \(A'\)},
                  {"B", \(B'\)}
                  }], ")"}], "|"}], "="}], TraditionalForm]]],
  "1, and if ",
  Cell[BoxData[
      \(TraditionalForm\`z\_\(\(\[LeftDoubleBracket]\)\(1\)\(\
\[RightDoubleBracket]\)\), \ 
      z\_\(\(\[LeftDoubleBracket]\)\(2\)\(\[RightDoubleBracket]\)\), \ 
      and\ z\_\(\(\[LeftDoubleBracket]\)\(3\)\(\[RightDoubleBracket]\)\)\)]],
  " are given by A/B, (A+\[ImaginaryI] A')/(B+\[ImaginaryI] B'), and (A-\
\[ImaginaryI]A')/(B-\[ImaginaryI]B') respectively, then the diameter of the \
circle through ",
  Cell[BoxData[
      \(TraditionalForm\`z\_\(\(\[LeftDoubleBracket]\)\(1\)\(\
\[RightDoubleBracket]\)\), \ 
      z\_\(\(\[LeftDoubleBracket]\)\(2\)\(\[RightDoubleBracket]\)\), \ 
      and\ z\_\(\(\[LeftDoubleBracket]\)\(3\)\(\[RightDoubleBracket]\)\)\)]],
  " is 1/Re[B*Conjugate[B']]. Here we give the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " code for a computer-algebra verification of the claim. "
}], "Subsubtitle"],

Cell[BoxData[
    \(mytimes[a_, b_] := {a[\([1]\)]\ b[\([1]\)] - a[\([2]\)]\ b[\([2]\)], 
        a[\([1]\)] b[\([2]\)] + a[\([2]\)]\ b[\([1]\)]}\)], "Input",
  InitializationCell->True],

Cell[TextData[{
  "This implements complex multiplication as an operation on pairs of real \
numbers. We have to do it this way because only thus can we get ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " to \"think about\" the real and imaginary parts of complex numbers as \
algebraic formal objects in their own right. "
}], "Text"],

Cell[BoxData[
    \(midpt[z1_, z2_] := \((z1 + z2)\)/2\)], "Input",
  InitializationCell->True],

Cell[BoxData[
    \(iit[x_] := {\(-x[\([2]\)]\), x[\([1]\)]}\)], "Input",
  InitializationCell->True],

Cell["\<\
This implements multiplication by i, and the next function implements the \
complex conjugate. \
\>", "Text"],

Cell[BoxData[
    \(myconjug[a_] := {a[\([1]\)], \(-a[\([2]\)]\)}\)], "Input",
  InitializationCell->True],

Cell[BoxData[
    \(ctral[z1_, z2_, z3_] := 
      Module[{d1, d2, c1, c2, s, t, s1, t1, u}, \[IndentingNewLine]d1 = 
          iit[z2 - z1]; d2 = iit[z3 - z1]; 
        c1 = midpt[z1, z2]; \[IndentingNewLine]c2 = 
          midpt[z1, 
            z3]; \[IndentingNewLine]u = \(Solve[{c1 + s\ d1 \[Equal] 
                  c2 + t\ d2}, {s, t}]\)[\([1]\)]; \[IndentingNewLine]c1 + 
            s\ d1 /. u]\)], "Input",
  InitializationCell->True],

Cell["\<\
The function ctral finds the center of the circle through z1, z2, and z3. The \
function below, mydiv, implements complex division as an operation on pairs \
of real numbers. \
\>", "Text"],

Cell[BoxData[
    \(mydiv[a_, b_] := 
      mytimes[a, myconjug[b]]/\(mytimes[b, myconjug[b]]\)[\([1]\)]\)], "Input",\

  InitializationCell->True],

Cell["\<\
The function radl computes the  square radius of the circle through z1, z2, \
and z3. \
\>", "Text"],

Cell[BoxData[
    \(radl[z1_, z2_, z3_] := 
      Module[{c}, 
        c = ctral[z1, z2, z3]; \((z1 - c)\) . \((z1 - c)\)]\)], "Input",
  InitializationCell->True],

Cell[CellGroupData[{

Cell[BoxData[
    \(mytimes[{2, 1}, {3, 2}]\)], "Input"],

Cell[BoxData[
    \({4, 7}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(mydiv[{2, 1}, {3, 2}]\)], "Input"],

Cell[BoxData[
    \({8\/13, \(-\(1\/13\)\)}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(\((2 + I)\)/\((3 + 2\ I)\) \[Equal] \((8 - I)\)/13\)], "Input"],

Cell[BoxData[
    \(True\)], "Output"]
}, Open  ]],

Cell["\<\
(It never hurts to check!) The function myabs, below, implements the complex \
absolute value as an operation on pairs of real numbers. Myreal and myimag \
implement the real and imaginary parts as functions on pairs of real numbers. \
\
\>", "Text"],

Cell[BoxData[
    \(myabs[a_] := Sqrt[\(mytimes[a, myconjug[a]]\)[\([1]\)]]\)], "Input",
  InitializationCell->True],

Cell[CellGroupData[{

Cell[BoxData[
    \(myabs[{3, 4}]\)], "Input"],

Cell[BoxData[
    \(5\)], "Output"]
}, Open  ]],

Cell[BoxData[{
    \(\(myreal[a_] := a[\([1]\)];\)\), "\n", 
    \(myimag[a_] := a[\([2]\)]\)}], "Input",
  InitializationCell->True],

Cell["\<\
We now declare A=an={an1,an2}, B={bn1,bn2}, and B'={bm1,bm2}. A' is not free, \
but is determined by the requirement on the determinant. 
\
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(an = {an1, an2}; bn = {bn1, bn2}; bm = {bm1, bm2}\)], "Input",
  InitializationCell->True],

Cell[BoxData[
    \({bm1, bm2}\)], "Output"]
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Cell[CellGroupData[{

Cell[BoxData[
    \(am = mydiv[mytimes[an, bm] + {1, 0}, bn]\)], "Input",
  InitializationCell->True],

Cell[BoxData[
    \({\(\((1 + an1\ bm1 - an2\ bm2)\)\ bn1 + \((an2\ bm1 + an1\ bm2)\)\ \
bn2\)\/\(bn1\^2 + bn2\^2\), \(\((an2\ bm1 + an1\ bm2)\)\ bn1 - \((1 + an1\ \
bm1 - an2\ bm2)\)\ bn2\)\/\(bn1\^2 + bn2\^2\)}\)], "Output"]
}, Open  ]],

Cell["\<\
And now we compute z1, z2, and z3. It gets messy for z2 and z3 so the \
displays have been suppressed. \
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(z2 = mydiv[an + iit[am], bn + iit[bm]];\)\), "\[IndentingNewLine]", 
    \(\(z3 = mydiv[an - iit[am], bn - iit[bm]];\)\), "\[IndentingNewLine]", 
    \(z1 = mydiv[an, bn]\)}], "Input",
  InitializationCell->True],

Cell[BoxData[
    \({\(an1\ bn1 + an2\ bn2\)\/\(bn1\^2 + bn2\^2\), \(an2\ bn1 - an1\ \
bn2\)\/\(bn1\^2 + bn2\^2\)}\)], "Output"]
}, Open  ]],

Cell["\<\
And now we compute the radius, as a formal algebraic object. It's a mess too. \
Display it at your peril. But the simplified version is exactly what it was \
claimed to be. \
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(u = radl[z1, z2, z3]; uu = Simplify[u]\)], "Input",
  InitializationCell->True],

Cell[BoxData[
    \(1\/\(4\ \((bm1\ bn1 + bm2\ bn2)\)\^2\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(u\)], "Input"],

Cell[BoxData[
    \(\((\(an2\ bn1 - an1\ bn2\)\/\(bn1\^2 + bn2\^2\) + \((\((\(-bm2\)\ bn1 + \
bn1\^2 + bm1\ bn2 + bn2\^2)\)\ \((\(-\(\(an1\ bn1 + an2\ bn2\)\/\(bn1\^2 + \
bn2\^2\)\)\) + \((\(-\((\(-bm1\) - bn2)\)\)\ \((an2 + \((\((1 + an1\ bm1 - \
an2\ bm2)\)\ bn1 + \((an2\ bm1 + an1\ bm2)\)\ bn2)\)/\((bn1\^2 + bn2\^2)\))\) \
+ \((\(-bm2\) + bn1)\)\ \((an1 - \((\((an2\ bm1 + an1\ bm2)\)\ bn1 - \((1 + \
an1\ bm1 - an2\ bm2)\)\ bn2)\)/\((bn1\^2 + bn2\^2)\))\))\)/\((\((\(-bm2\) + \
bn1)\)\^2 - \((\(-bm1\) - bn2)\)\ \((bm1 + bn2)\))\))\))\)/\((2\ \((bm1\ bn1 \
+ bm2\ bn2)\))\) + 1\/2\ \((\(-\(\(an2\ bn1 - an1\ bn2\)\/\(bn1\^2 + bn2\^2\)\
\)\) - \((\((\(-bm2\) + bn1)\)\ \((an2 + \(\(1\/\(bn1\^2 + bn2\^2\)\)\((\((1 \
+ an1\ bm1 - an2\ bm2)\)\ bn1 + \((an2\ bm1 + an1\ bm2)\)\ bn2)\)\))\) + \
\((\(-bm1\) - bn2)\)\ \((an1 - \(\(1\/\(bn1\^2 + bn2\^2\)\)\((\((an2\ bm1 + \
an1\ bm2)\)\ bn1 - \((1 + an1\ bm1 - an2\ bm2)\)\ bn2)\)\))\))\)/\((\((\(-bm2\
\) + bn1)\)\^2 - \((\(-bm1\) - bn2)\)\ \((bm1 + bn2)\))\))\))\)\^2 + \((\(an1\
\ bn1 + an2\ bn2\)\/\(bn1\^2 + bn2\^2\) + 1\/2\ \((\(-\(\(an1\ bn1 + an2\ bn2\
\)\/\(bn1\^2 + bn2\^2\)\)\) - \((\(-\((\(-bm1\) - bn2)\)\)\ \((an2 + \
\(\(1\/\(bn1\^2 + bn2\^2\)\)\((\((1 + an1\ bm1 - an2\ bm2)\)\ bn1 + \((an2\ \
bm1 + an1\ bm2)\)\ bn2)\)\))\) + \((\(-bm2\) + bn1)\)\ \((an1 - \
\(\(1\/\(bn1\^2 + bn2\^2\)\)\((\((an2\ bm1 + an1\ bm2)\)\ bn1 - \((1 + an1\ \
bm1 - an2\ bm2)\)\ bn2)\)\))\))\)/\((\((\(-bm2\) + bn1)\)\^2 - \((\(-bm1\) - \
bn2)\)\ \((bm1 + bn2)\))\))\) + \((\((\(-bm2\)\ bn1 + bn1\^2 + bm1\ bn2 + bn2\
\^2)\)\ \((\(an2\ bn1 - an1\ bn2\)\/\(bn1\^2 + bn2\^2\) - \((\((\(-bm2\) + \
bn1)\)\ \((an2 + \((\((1 + an1\ bm1 - an2\ bm2)\)\ bn1 + \((an2\ bm1 + an1\ \
bm2)\)\ bn2)\)/\((bn1\^2 + bn2\^2)\))\) + \((\(-bm1\) - bn2)\)\ \((an1 - \
\((\((an2\ bm1 + an1\ bm2)\)\ bn1 - \((1 + an1\ bm1 - an2\ bm2)\)\ \
bn2)\)/\((bn1\^2 + bn2\^2)\))\))\)/\((\((\(-bm2\) + bn1)\)\^2 - \((\(-bm1\) - \
bn2)\)\ \((bm1 + bn2)\))\))\))\)/\((2\ \((bm1\ bn1 + bm2\ bn2)\))\))\)\^2\)], \
"Output"]
}, Open  ]]
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(*******************************************************************
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*******************************************************************)