Remarks Concerning Certain Homework Problems
\textit{ $\C$  2010, Prof. George K. Francis, Mathematics Department,
University of Illinois}

\begin{document}
\maketitle

\section{Regarding Homework Problem 8.2.14}
Unhappily, it appears that no-one in the class got this problem right. Almost
everyone ignored one part or another of the hypothesis, and did whatever was
convenient about those absolute value signs. Many of you
ignored the class notes for W12 which begins with the
explicit statement "This lesson replaces Theorem 8.11 in the book." Hvidsten's
exposition is based on prior work which we did not do in this class.
Moreover, here the difference between Hvidsten's cross ratio $(a,b,c,d)$ and
our cross ratio $CR(a,b,c,d)$ really matters.

The formula $\frac{a-b}{a-d}\frac{c-d}{c-b}$ is entirely unambiguous. But
there is a choice whether to write this as $CR(a,b,c,d)$ as we did, or
as $(a,c,b,d)$ as Hvidsten, p. 317 chose to do it.

Our choice is again explained after the solution.
And a guess about why Hvidsten made his choise is given even further down.
If your journal has a mixture of these, you should make the appropriate
corrections as you use the journal to study for the final.

At any event, on a test, it is wise to make it clear exactly what you
mean by the abbreviation $CR(a,b,c,d)$. And watch the subscripts, if
there are some.

\section{Complete solution in utter detail.}

I will first give a correct solution. And then discuss how your reviewing
this section of the course can profit from the above errors.

\subsection{Did you use the hypothesis?}
The hypothesis, that $z_1$ lies \textit{between} $z_0, z_2$ cannot be
ignored. Consider a counterexample:

If $z_0 = 0, z_1 = 2, z_2 = 1$ then even for Euclidean distance
$1 = d(0,1) \ne 3 = d(0,2) + d(2,1).$

\subsection{Normalize the problem.}
In the notes on this section we repeatedly move the circline bearing
the 3 points to the diameter by an isometry (so the distances do not
change). So move $z_0 \mapsto 0$ by a hyperbolic translation. If
if the diameter happens to be at a slant, a rotation (about the origin)
will move it to the real diameter, namely so that
$z_1 \mapsto r, z_2 \mapsto t$ and now the hypothesis $0 < r < t < 1$
comes into play.

\subsection{Getting rid of the absolute values.}
For $0 < r < 1$, we evaluated the crossratio and found that
$d_H(0,r) = log \frac{1+r}{1-r}$. The absolute values bars disappear
because $1 < \frac{1+r}{1-r}$. Since we calculated this one in the
notes, let us calculate the second distance:

\begin{eqnarray*}
d_H(r,t)&=& | log CR(r,-1,t,+1)| \\
&=& | log \frac{r+1}{r-1}\frac{t-1}{t+1}| \\
&=& | log \frac{1+r}{1-r}\frac{1-t}{1+t}| \\
&=& - log \frac{1+r}{1-r}\frac{1-t}{1+t} \\
&=& + log \frac{1-r}{1+r}\frac{1+t}{1-t} \\
\end{eqnarray*}

If you don't get that minus sign in line 4 right, you would have
a reciprocal in the last line and the laws of logarithms wouldn't
work as they need to thus:

\begin{eqnarray*}
d_H(0,r) + d_H(r,t) &=& log \frac{1+r}{1-r} + log \frac{1-r}{1+r}\frac{1+t}{1-t} \\
&=& log \frac{1+r}{1-r}\frac{1-r}{1+r}\frac{1+t}{1-t} \\
&=& log \frac{1+t}{1-t}  \\
&=& d_H(0,t).  \\
\end{eqnarray*}

And that minus sign in the 4th line itself requires a simple proof.
Namely, if $0 < r < t < 1$
then which is bigger? You might guess (hope?) that

\begin{eqnarray*}
\frac{1+r}{1-r} & < ?  & \frac{1+t}{1-t} \\
1 + r - t - tr & < ?  & 1 + t - r - rt \\
2r & < !  & 2t \\
\end{eqnarray*}

and so it is. Now this problem has been solved in the way we studied this
material.

\section{So, what about using the textbook?}
I explained at some length that we used a different definition of the
cross ratio symbol than Hvidsten does. For the final you'll need only understand
and use the one we developed in class. Read your class notes! And you
can safely ignore what follows here. However, in all fairness, I will go over
the differences. There are no errors in either approach. I think Hvidsten's
is more likely to lead to confusion than mine. But using both at once is
guaranteed to get confusing.

\subsection{The mnemonic}
For present purposes, I will use $CR(a,b,c,d)$ and $HV(a,b,c,d)$.
If $f(z) = CR(z,z_0,z_1,z_\infty) = \frac{z-z_0}{z-z_\infty}\frac{z_1 - z_\infty}{z_1-z_0}$
takes $z_0 \mapsto 0, z_1 \mapsto 1, z_\infty \mapsto \infty$ then
Hvidsten chose the order of his cross so that the same function would be
written  $f(z) = HV(z,z_1,z_0,z_\infty)$. To avoid this transposition
I chose the particular order we did.

Moreover, in Hvidsten's distance formula, the one we did \textit{not}
adopt, he uses a Ruler on the Poincare Line $\ell_{z_0, z_1}$
which orients it from $z_1$ to $z_0$, while our definition keeps the
natural order from the first mentioned point to the second mentioned point.
(I hope you noticed the capital R. This ruler is non-Euclidean, but it still
has the additive property on Lines. That's what this problem is about.)

The reason there is no contradiction is absolute
values around the logarithms.  If you compare the two cross ratios
they differ by being reciprocals. But
$| log \frac{n}{d} | = | - log \frac{d}{n} | = | log \frac{d}{n} |.$

Finally, you can't just ignore the absolute values when you want to
take the "sum of two logarithms" to the "logarithm of the product".
You have to make sure that you're working with positive logarithms
i.e. that they are logarithms of real numbers \textit{bigger} than $1$.
And that's were the order of the three points matters. This is stated
in Hvidsten's proof of Theorem 8.11, but most students miss that
hypothesis and get it wrong on an application.

To summarize, there are two issues here. First, you don't take
logarithms of complex numbers (though it can be defined, we don't
take $log (1+i)$, for example.) So, we make sure that the
cross ratios we take the logarithm of
come out to be real. And that is, if and only if the four
points lie on a circline.

Second, when measuring a distance we
want the logarithms to be positive so we can ignored those absolute
value signs. For that the real number you take the logarithm of needs
to be greater than 1. That's where the order of $z_0,z_1,z_2$ matters.
Hvidsten reverses the ruler, I think, so that  for $0 < r < 1$ we have
$d_H(0,r) = log \frac{1+r}{1-r} = log CR(0,+1,r,-1) = HV(0,r,1,-1)$
comes out directly, whereas we had to really use the absolute values
or distance would come out negative.

The moral of this story is that when there is a deviation from the
approach taken in the textbook, try not to serve two masters, something
won't come out right.