Exercises on Moebius Transformations
revised 21apr11\\
\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 problems 8.2.1-8.2.15. There is
a great deal of geometry hidden in these problems, which is obscured by
their brevity. The elaborations here is intended to help you use the
exercises to understand the facts and methods. A "solution" to a problem
should be more of a mini essay on what the problem means, than just cranking
out what looks like an answer to a direct question. In this spirit we proceed.

\section{Regarding Homework Problems for Chapter 8}
\subsection{Exercises on Hvidsten p325}
\begin{enumerate}
\item Ex 8.2.1 "Show that the $w=e^{i\phi}+b$ form a group, the group of
rigid Euclidean motiosn (a.k.a. Euclid's congruences) of the plane."
\item Remember, to show that a
subset of the group of all bijections 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.
\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 Ex 8.2.2
"Show that the equation of a line can be given as
$\mathfrak{Im}(\alpha z + \beta) =0$." While this requires only some
arithmetic applied to the definitions, it has a deeper meaning. Note
that $\mathfrak{Im}(w)=0$ is the same thing as saying
$\mathfrak{Re}(w)\in \mathbb{R}.$  Since a straight line is a special
case of a circline, this 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 $\mathfrak{Im}CR(z,z_0,z_1,z_\infty)=0$."

\item Ex 8.2.3 in my edition of the  text  has a misprint here.
The notes show that the equation for 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, since the 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. Incidentally, if the three points are on a 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.

\item Supplementary Question. Write $f(z)=az+b$ as a cross ratio. This
should be a one-liner.

\item As Hvidsten says: 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.

\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.

\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 Ex 8.2.8 Apropos the earlier comment, this asks you prove that a
particular subset is a subgroup. You'll need to check 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?)
By "$f$ fixes or preserves the unit circle"
we mean that if $|z|=1$ then $|f(z)|=1$ as well. In our current
edition of the notes, this exercise is solved there.

\item 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 does suffice to prove it
just for this case? Don't forget to answer this.)
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.

\item Ex 8.2.10 has a misprint and it also changes the form of
the isometry to $w' = \frac{z-\alpha}{1- \bar{\alpha}z}$ by multiplying
the original form $w = \frac{z-\alpha}{\bar{\alpha}z -1 }$ by $-1$.
In effect, it involves a rotation by $180^o$ and might be
written $w'= e^{i\pi}\frac{z-\alpha}{\bar{\alpha}z -1 }$.

Setting $w=z$
and cross multiplying yields a quadratic equation
$\bar{\alpha}z^2 - z + \alpha$ which has two solutions
$z_i=\frac{1\pm \sqrt{1- |\alpha|^2}}{\bar{\alpha}}$, unless $\alpha = 0$.
But where are these two points? Hvidsten cleverly changes the problem (note
the denominator). He considers the case that $\theta = \pi$ in the isometry.
Now the quadratic equations is much simpler, and the solutions is
$z_i = \pm \frac{\alpha}{|\alpha|}$ which obviously lies \textit{on},
not \textit{in}
the unit circle. So it is not a Point in the hyperbolic plane as
modelled by the inside of the unit disk. Note that capital P, do you
remember why we do that? In other words, the isometry
$T_{-\alpha}(z) := \frac{z-\alpha}{1-\bar{\alpha}z}$ leaves
no Point fixed in the
Hyperbolic plane, and takes $\alpha$ to $0$.  Incidentally,
$T_{-\alpha}(0) = -\alpha$.  (I have put a - sign into the subscript
for reasons that will become clear. This material is supplementary to
Hvidsten and needs to be understood.)

\item  These transformations have inverses of the same kind:
\begin{eqnarray*}
T_{-\alpha}^{-1}(z) &=& \frac{z+\alpha}{1+\bar{\alpha}z} \\
& = &T_{\alpha}(z) \\
T_{\alpha}^{-1} & = & T_{-\alpha} \\
\end{eqnarray*}

These transformations are also
are closed under composition in the sense that
$T_\beta T_\alpha = T_\gamma$, for a unique $\gamma$ given both
$\alpha$ and $\beta$. It is an instructive exercise to calculate $\gamma$,
but $\gamma$ does not have a deep geometrical meaning beyond its
existence. More interesting is a
composition like  $T_{\beta} T_{-\alpha}$, which takes $\alpha$ to
$\beta$ and which we write $T_{\alpha, \beta} := T_{\beta} T_{-\alpha}$
for reasons of its similarity to the analogous Euclidean translations by
vectors.
Note that these obey the triangle rule of vectors:
$T_{\beta, \gamma} \circ T_{\alpha, \beta} = T_{\alpha, \gamma}$ .
To verify this painlessly,
note that $T_{\alpha, \beta} = T_{\beta} T_{\alpha}^{-1}$ .
So these
\textit{hyperbolic translations} really do behave very much like
vectors in Euclidean geometry.