Z2 revised 4jul11, 8mar13
\section{What you should study now.}
Hvidsten Ch8, $\S 8.1$, and $\S 8.2$. It is not necessary to refer to
previous material in Hivdsten. These notes, though terse, are self-contained.
So it is possible for you to work directly from Hvidsten, correct the
(few) typos and errors, and solve the assigned problems (8.2.1--15).
However, you will
gain a much deeper understanding of "what's going one here" by following
these lessons. Here is advice on how to solve the problems in a more
meaningful way. And you will proceed in a more leisurely manner.
\section{Introduction}
By a \textit{transformation} of the plane we shall mean a bijection $w=f(z)$
of the complex numbers to themselves. This is a simple generalization from
functions on the real numbers to the complex number. Being bijective, we
also require that $f$ be one-to-one and onto.
Lesson on Moebius Transformations
\textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle
\section{What you should study now.}
Hvidsten Ch8, $\S 8.1$, and $\S 8.2$. It is not necessary to refer to
previous material in Hivdsten. These notes, though terse, are self-contained.
So it is possible for you to work directly from Hvidsten, correct the
(few) typos and errors, and solve the assigned problems (8.2.1--15).
However, you will
gain a much deeper understanding of "what's going one here" by following
these lessons. Here is advice on how to solve the problems in a more
meaningful way. And you will proceed in a more leisurely manner.
\section{Introduction}
By a \textit{transformation} of the plane we shall mean a bijection $w=f(z)$
of the complex numbers to themselves. This is a simple generalization from
functions on the real numbers to the complex number. Being bijective, we
also require that $f$ be one-to-one and onto.
Question 1.
Show that $f(z)=z^2$ is onto (surjective) but no 1:1 (injective). Hint:
For the former you must find at least one solution $z$
to the equation $w=z^2$ for \textit{every} $w$. For the latter it suffices to
find two different solutions for \textit{one} particular value of $w$.
Because each complex number is the name of a point in the plane and vice
versa, we can interpret a bijection of the complex numbers as a
\textit{point transformation of the plane.}
That is, each point is transformed into a
unique other point (which may or may not be itself.) Here we shall be
concerned only with functions which are very simple algebraically but
having very rich geometrical properties.
\section{Similitudes.}
A \textit{similitudes} of the plane is the transformation of every
point $z$ in the plane to another point $w$ obeying this
\[ \mbox{ linear equation: } w = az + b, \mbox{ where }a,b \in \mathbb{C}
\mbox{ and } a\ne 0.\]
The totality of all similitudes form a \texit{group} under composition
for the following reasons (MA347).
\begin{eqnarray*}
\mbox{closure under composition:} \ a(pz + q) + b &=& (ap)z + (aq + b) \\
\mbox{inverse of a similitude:} \ w = az + b &\implies& z = (\frac{1}{a})w + (\frac{-b}{a}) \\
\end{eqnarray*}
Observe that we carefully place parentheses so that $z$ can be seen to be
a linear polynomial of $w$. The only thing to verify is that $ap \ne 0$
provided neither $a$ nor $p$ is, and that $\frac{1}{a} \ne 0$. This is not
an entirely trivial observation. We shall see many instances in these
lessons where we claim that the compositions of two kinds of transformations
is again of the same kind. This means that the composition must be shown
to have the same \textit{ form} as the form that defines the kind of
transformation we are talking about.
Thus the identity function, written $\iota(z)=z$ for all $z$ is a
similitude because $\iota(z) = 1 z + 0 $.
Question 2.
Show that the composition of a similitude with its inverse is the
identity.
Composition of three transformation is always associative, no matter what
kind of a transformation they are. So, now we have shown that the the
traditional four conditions for a group are satisfied for similitudes.
Study this proof carefully. There are several more kinds of transformations
which need verified that they are groups, either in your Journal, the
homework, or both.
\subsection{Anatomy of a Similitude}
Since we can write $a = re^{i\theta}$ we see that
a similitude, $f(z)= re^{i\theta} z + b $, is just the
composition of three familiar transformations, $f=f_3 \circ f_2 \circ f_1$,
where:
\begin{itemize}
\item The \textit{rotation} $ f_1(z) = e^{i\theta}z $
\item is followed by the \textit{ dilation } $ f_2(z) = rz$
\item which is followed by the \textit{ translation } $ f_3(z)= z + b$.
\end{itemize}
The functional notation is often more conveniently replaced by
the \textit{ maps-to} notation, as in
\[ z \mapsto e^{i\theta}z \mapsto r(e^{i\theta}z) \mapsto (re^{i\theta}z) + b \]
\subsection{The Euclidean Group}
For a similitude to be a \textit{Euclidean transformation} there must
not by any \textit{dilation} (stretching or shrinking}.
Such similitudes have the form $f(z) = e^{i\theta}z+b $.
To show that the Euclidean
transformations also form a group we simply follow the above strategy.
Note that
a Euclidean transformation is just a similitude $w=az+b$ with $|a|=1$.
Noting that $|a|= |p| = 1 \implies |ap|=|a||p|=1$, we have closure under
composition. The composition of two Euclidean transformations is not
just another similitude, it's a Euclidean one .
Question 3.
Show that the inverses of a Euclidean similitude is again Euclidean.
This should be the last time we refer to these transformations as "Euclidean
similitudes" for the same reason that we don't call squares
"rhombic rectangles" or "rectangular rhombi", although both would be
correct. Their classical term is Eucliean \textit{ Motions} in the plane,
and, we shall see later, they are most properly called
\textit{Euclidean isometries}.
\subsubsection{Canonical Form for the Euclidean Group}
Sooner or later, you will have occasion to make use of the fact
that the rotation angle $\theta$ and the translation vector $b$
of a Euclidean transformation completely specifies the transformation.
\textbf{Theorem: }
If $ (\forall z)( e^{i\alpha}z + a = e^{i\beta}z + b)$ then
$ (\alpha = \beta) \and (a = b) $.
\textbf{Proof: } Since the equation holds for all $z$, it holds for
$z=0$. From this it follows that $a=b$. Subtract the constant from
both sides, and substitute $z=1$ to get that $ \alpha = \beta \ mod(2\pi)$.
Study this proof strategy because we shall see it several more times, in
the lessons, homework, and, of course, on tests.
\subsection{Rotations}
Recall from an earlier lesson that the rotation of the complex plane
by an angle of $\theta$ about a point $c \ne 0$ can be considered
as the composition of three Euclidean tranformations:
\[ z \mapsto z-c \mapsto e^{i\theta}(z-c) \mapsto e^{i\theta}(z-c) +c .\]
In words, first translate every point so that $c$ is moved to the
origin. Then rotate the plane about the origin, and finally, translate
the plane so that the origin moves back to the original center of
rotation.
First note that this too is a Euclidean transformation, and for two
different reasons. First, because the Euclidean transformations are
closed under composition. But we can see it directly, by rewriting
\[ e^{i\theta}(z-c) +c = e^{i\theta}z + (1-e^{i\theta})c,\]
which is a rotation about the origin followed by a translation, which
is the canonical form of a Euclidean transformation.
Question 4.
If $\tau(z)=z+c$ and $\rho(z) = e^{i\theta}z $ then what is the
transformation $f = \tau \circ \rho \circ \tau^{-1}$ ?
The composition $g \circ f \circ g^{-1} $ is called the
\textit{ conjugation} of the transformation $f$ by $g$. It is
important to distinguish the two uses of the word "conjugate" in
the algebra of complex numbers. Recall that the \textit{ conjugate }
of $z = x +iy$ is defined as $\bar{z} = x-iy$.
Question 5.
Write the conjugate of $z = re^{i\theta}$ in polar form.
Question 6.
If $f(z)= \frac{1}{z}$, the reciprocal function, and $g(z)= z+b$ is a
translation, then what is the conjugate of $f$ by $g$? What is the
congugate of $g$ by $f$?
\section{Reciprocals and Inversion}
While a similitude taks a complex number to a complex number, the reciprocal
of a complex number has a problem at $z=0$. Unlike for real numbers, for
complex numbers we can invent a single \textit{point at infinity}, and
extent the complex plane to the \textit{ Riemann sphere},
denoted by $\hat\mathbb{C} = \mathbb{C} + \infty$. Think of the north pole
on the earth. To some extent complex arithmetic also extends as follows:
\begin{eqnarray*}
\frac{1}{0} &=& \infty \\
\frac{1}{\infty} &=& 0 \\
\infty \pm b &=& \infty \\
\infty b &=& \infty \\
\frac{a \infty + b}{c \infty + d}&=& \frac{a}{c} \\
\end{eqnarray*}
The way to understand this strange arithmetic is, in each case, to
subsitute $\frac{1}{z}$ for each occurence $\infty$ and take the
$ \lim_{z -> 0}$. It is instructive to illustrate this method for
the last example.
\begin{eqnarray*}
\frac{ a/z + b}{c/z + d} &=& \frac{a + zb}{c + zd} \\
\lim_{z->0} \frac{ a + zb}{c + zd} &=& \frac{a}{c}\\
\end{eqnarray*}
In particular, the function $f(z) = \frac{1}{z} $ is called the
\textit{inversion of the complex plane}. Note that
\begin{eqnarray*}
f(0)&=&\infty \\
f(\infty)&=& 0 \\
f(e^{i\theta}) &=& e^{-i\theta} \\
\end{eqnarray*}
where the last equations says that the unit circle remains fixed as
a set, even though the points on the unit circle themselves are not
fixed, except $\pm 1 $.
\section{Moebius Transformation}
The final transformation we define in this section is called a
\textit{fractional-linear} transformation, or more commonly, a
\textit{Moebius transformation}:
\begin{eqnarray*}
\mu(z;a,b,c,d))&=& \frac{az+b}{cz+d} \ where \ ad-bc = 1 \\
\end{eqnarray*}
\subsection{Our Definition}
Note carefully that we use the customary definition, not the definition
in Hvidsten, who merely insists that $ad-bc\ne 0$. It is important to
understand the difference of these two definitions, especially how they
are used.
Of course $ad-bc = 1$ implies that $ad-bc \ne 0$. But even if we merely
say that $ad-bc \ne 0$, we may rewrite the fraction so that the function
assumes the seemingly more restrictive form without changing the function.
For, suppose initially that $\delta := \sqrt{ad - bc} \ne 1 $ but still
non-zero. Recall that every non-zero complex numbers has (two) squareroots.
You can use either.
Then divide numerator and denominator by this number to obtain a new
set of coefficients.
\begin{eqnarray*}
\frac{az+b}{cz+d} &= &\frac{(a/ \delta )z+(b/\delta)}{(c/ \delta)z+(d/\delta)}
&=& \frac{a'z+b'}{c'z+d'} \ and\ a'd' - b'c' = \frac{ad-bc}{\delta^2}=1. \\
\end{eqnarray*}
This is not rocket science, you've known this idea since fourth grade.
There, you had to get used to the idea that $ \frac{2}{4}=\frac{1}{2}$
even though $2 \ne 1$ and $4 \ne 2$. In the first case we define the
fraction as the ratio of two whole numbers, with the denominator not
being zero. In the second case we express the same fraction in "lowest
terms" as they said in 4th grade. In college (MA347) you learned to
say that numerator and denominator are \textit{relatively prime}.
A (positive) fraction has unique relatively prime numerator and denominator.
So too, we say that the expression for the Moebius transformation is
\textit{ canonical} if $ad-bc=1$.
\subsection{The Moebius Group}
Note that, for $c \ne 0$, $\mu(\frac{-d}{c};a,b,c,d) = \infty $. Moreover,
$\mu(\infty;a,b,c,d) = \frac{a}{c} $. So some finite point goes to
infinity, and infinity goes to a finite point. Therefore, a Moebius
transformation is in reality a function not of the complex plane
$\mathbb{C}$ but of the Riemann sphere $\hat\mathbb{C}$.
That the set of all Moebius transformation of the Riemann sphere
form a group is just algebra again.
Question 7.
Show that the composition of two Moebius transformations is again a
Moebius transformation.
Question 8.
Prove that the inverse of a Moebius tranformation is again a Moebius
transformation by solving $w=\frac{az+b}{cz+d}$ for
$z =\frac{a'w+b'}{c'w +d'}$. Show that $a'=d, b'= -b, c'=-c, d'=a$.
\subsection{Aside on Matrices}
If you have had matrix algebra you will recognize the last exercises
as the formula
\[\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right)^{-1}=
\left(
\begin{array}{cc}
d & -b \\
-c & a \\
\end{array}
\right)
\]
for the inverse of a linear transformation in the plane. And if
you composed the $g(z)=\mu(z;a,b,c,d)$ with $f(z)=\mu(z;p,q,r,s)$
for the exercise on composition, you may have noticed that you
just multiplied the two corresponding matrices.
Indeed, the Moebius group may be represented by the group
of complex 2x2 matrices with unit determinant,
the \textit{ Special Linear Group }
$SU(2,\mathbb{C})$. But we do not require linear algebra as
a prerequisite to this course, so we shall not pursue this subject
further.
\section{The Anatomy of a Moebius Transformation}
Note that when $c=0$ the Moebius transformation
$f(z)=\frac{az+b}{cz+d}$ is just a similitude, with
dilation factor $|a/d|$, rotation angle $arg(a/d)$, and
translation vectore $b/d$.
We close this lesson with Hvidsten's first theorem which says that
every Moebius transformation is either a similitude ($c=0$), or
the composition of a single inversions with more similitudes.
Algebraically, we said nothing other that,
\[\frac{az+b}{cz+d} = \frac{a}{c} - \frac{ad-bc}{c^2}\frac{1}{z+ d/c}.\]
One immediate consequence of this fact is that every Moebius
transformation is a bijection of the Riemann sphere, because
each component transformation, dilation, rotation, translation and
inversion is trivially a bijection. Write this argument out
(in full detail) and in your own words, into your Journal.
\subsection{Verification versus Discovery}
Hvidsten rewrites a Moebius tansformations as a succession of rotations,
dilations, and inversions. All the reader needs to do is check that the
equation is correct. If the transformation is already a similitude, then
there is no inversion. By writing $az+b = re^{i\theta}z + b$, we see how
a similitude is the composition of a rotation, followed by a dilation,
followed by a translation. But how would you discover such
a formula in the general case? Study the following steps to see the trick
of "cancelling $cz+d$ in the numerator and denominator".
Consider this way of proving Hvidsten's first theorem (HP8.1):
\textbf{Theorem: }
Every proper ($c\ne0$) Moebius transformation is the composition of two similitudes with
an inversion between them.
\textbf{Proof:}
\begin{eqnarray*}
\frac{az+b}{cz+d}\frac{c}{c} &=& \frac{acz+bc}{c(cz+d)} \\
&=& \frac{acz+ad + (bc-ad)}{c(cz+d)} \\
&=& \frac{a}{c}[\frac{cz+d}{cz+d}] + \frac{bc-ad}{c} (\frac{1}{cz+d}) \\
&=& \frac{a}{c} + \frac{bc-ad}{c} (\frac{1}{cz+d}) \\
&=& B + A (\frac{1}{cz+d}) \\
\mbox{where } A &=& \frac{bc-ad}{c} \\
\mbox{and } B &=& \frac{a}{c} \\
\end{eqnarray*}