Exercises on Moebius Transformations
revised 21apr11, 13jul11\\
\textit{ $\C$  2010, Prof. George K. Francis, Mathematics Department,
University of Illinois}

\begin{document}
\maketitle

\section{Introduction}
This lesson is a commentary on Hvidsten's \textit{Exercises 8.2.1-8.2.15}
on page 325. There is a great deal of geometry hidden in these problems, which
is obscured by the brevity of their statement, or even their solution.
The elaborations here is intended to help you use the exercises to understand
the facts and methods. A "solution" to a problem
should be a mini essay on what the problem means, and not consist of
just cranking out what looks like an answer to a direct question.
In this spirit we proceed.

\section{Discussion of the Exercises}
\subsection{Transformations which form Groups}

\begin{enumerate}
\item \textbf{Ex. 8.2.1} "Show that the $w=e^{i\phi}+b$ form a group, the group of
rigid Euclidean motions (a.k.a. Euclid's congruences) of the plane." This is
special case of the group of \textit{similitudes} treated extensively in the
lessons.
\item Remember, to show that a
subset of the group of all bijective functions of the extended plane is itself
a group, all you need to do is to show that
\begin{itemize}
\item \textit{Closure: }The composition of two such
function can be put into the same canonical form that defines the subset.
\item \textit{Inverse: } The inverse of such a function can be put
into the same canonical form.
\end{itemize}
Be sure you have this example in your Journal and are able to reproduce
the definitions and proof of this theorem on a test.

\item As Hvidsten says: \textbf{Ex 8.2.4 - 8.2.7} are steps in the proof that the set of Moebius
transformations also form a group. Much of this is in the notes. These
4 problems help you to organize this material in a coherent manner in your
Journal. As long as the actual calculations are short, a test question
may ask you to check that a given set of MTs form a group.
\begin{itemize}
\item Ex 8.2.4 this is the closure property and is pure high school algebra.
Substitute the result of the first transformation into the formula for the
second and rewrite in the standard form of Moebius transformation. Use our
definition, where $ad-bc=1$ is the condition on the coefficients. You'll be
happier for it.

\item Ex 8.2.5
As in our notes, the presence of the identity follows from
the presence of the inverse (8.2.6) and closure (8.2.4). However, an
immediate guess will solve this problem directly. But here there is an
easier solution by inspection.

\item  Ex 8.2.6 you'll need to solve $w=\frac{az+b}{cz+d}$
for $z =\frac{ez+f}{gz+h}$, by giving $e,f,g,h$ as functions of
$a,b,c,d$. But Hvidsten suggests a simpler approach. For this to hold,
you'll need to use his criterion $ad-bc\ne 0$. Recall why we can
assume either condition. These calculations belong into your Journal.

\item Ex 8.2.7 is done below. The answer is that transformations
always are associative: $(k \circ h) \circ g = k \circ (h \circ g)$.
All you need to remember is how to test that two functions are the same.
\end{itemize}

\item Solution to 8.2.7: An equation of functions holds provided the
equation holds for every value in the domain of the functions. So
applying LHS to an arbitrary complex number $z$ we see that
$(k\circ h)\circ g (z) = (k \circ h)(g(z)) = k(h(g(z)))$ which is
what you get if you plug $z$ into the RHS.

\item \textbf{Ex 8.2.8: } This exercise asks you prove that a particular subset
of transformations which have a particular geometric property,forms
a group. As we have seen, all you  need to check is closure and inverses.
It is unfortunate that Hvidsten uses the word "fix" instead of "preserve".
He does not mean that each point on the unit circle remains fixed under
the transformation. (That would mean it is the identity, right? since
there are at least 3 points on the circle.))
By "$f$ fixes or preserves the unit circle" we mean it maps every point
on the circle back to some (possibly different) point on the same circle.
Algebraically, if $|z|=1$ then $|f(z)|=1$ as well.

\end{enumerate}

\subsection{CircLine Geometry}
\begin{enumerate}

\item \textbf{ Ex 8.2.2 }
"Show that the equation of a line can be given as
$\mathfrak{Im}(\alpha z + \beta) =0$." While you can prove this with
just a little arithmetic applied to the definitions, it has a deeper meaning.
Note that for any complex number $w$,  $\mathfrak{Im}(w)=0$ is the same
thing as saying  $w \in \mathbb{R}.$

\item \textbf{Extension 8.2.2bis} Since a straight line is a special case of a circline, this problem
is a special case of the general theorem that the equation of a
circline through three points $z_0,z_1,z_\infty$is given by
$\mathfrak{Im}CR(z,z_0,z_1,z_\infty)=0$. Equivalently that
$CR(z,z_0,z_1,z_\infty) \in \mathbb{R}$.

\item \textbf{ Ex 8.2.3}
In my edition of the text  has a misprint here. The correct
equation of the unit circle is $|z|=1$.
"Find a Moebius transformation that takes the unit circle to the x-axis."
There are, of course, very many solutions. But there is a "canonical"
solution in terms of cross-ratios, if you think of the unit circle as
$\odot(1,i,-1)$. Use the theorem that an  MT defined by
$f(z)=CR(z,z_0,z_1,z_\infty)$ takes the circline $\odot(z_0,z_1,z_\infty)$
to the real-line.

\item \textbf{Extension 8.2.3bis}
Incidentally, if the three points are on any circle,
and in positive (counter-clockwise) order, then this MT takes the interior of
the circle to the upper halfplane, and the exterior to the lower halfplane.
\end{enumerate}

\subsection{The Hyperbolic Group}
The set of MTs which "fix" the unit circle form the group of isometries
in the hyperbolic plane, called the \textit{ Hyperbolic Group} for short.
The way this is established is exploit .
\begin{itemize}
\item Calculate the canonical form of every such MT.
\item Use one property of such an MT to define a hyperbolic line.
\item Exploit a property of hyperbolic lines so defined in terms of a cross ratio.
\item Define a ruler, in the sense of Birkhoff, for every line.
\item Verify that Birkhoff's Ruler Axiom is indeed verified.
\end{itemize}
The Exercises in the textbook serve this program.

\begin{enumerate}
\item \textbf{ Ex 8.2.9:} This is the \textit{ transitivity } of the Hyperbolic
Group.  By \textif{isometry} here Hvidsten means a
Moebius transformation which preserves the unit circle. In Prop.8.10
he proves that such a Moebius transformation must have the form
\begin{eqnarray*}
w = e^{i\theta}\frac{z-\alpha}{\bar{\alpha}z -1} & \mbox{where} &  |\alpha|<1\\
\end{eqnarray*}
So now the problem reduces to finding angle $\theta$ and point $\alpha$
which is moved to the origin under the isometry for given values for $z$ and
$w$. Hvidsten  he calls $z=P$ and $w=Q$.
But Hvidsten gives you a hint, by asking the question for the case
that $z$ remains arbitrary, but $w=0$. (Why is it enough to prove it just
for this case? Don't forget to answer this in your solution.)
Finally, for this special case the answer is now very easy,
namely, to solve
\begin{eqnarray*}
0 = e^{i\theta}\frac{P-\alpha}{\bar{\alpha}P -1} & \mbox{where} &  |\alpha|<1\\
\end{eqnarray*}
we can take $\theta = 0$, i.e. no rotation, and $\alpha = P$, which is
legitimate, because we are given that $|P| < 1$. The isometry is not unique,
but it is unique up to a rotation.
\end{enumerate}

[The remaining exercises will be discussed here in the future.]

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