## Lesson on Symmetric Points

Z4 7apr15

$\C$ 2010 2015, Prof. George K. Francis, Mathematics Department, University of Illinois. \begin{document} \maketitle \section{Introduction} In this (provisional) lesson we study the Symmetry Principles of Moebius Transformations we have used without proof in previous lessons. \textbf{Properties Moebius Transformations} \begin{itemize} \item Circles and lines (circlines) are mapped to circlines. \item Pairs of symmetric points to a circline are mapped to symmetric pairs. \item The orientation of a circline is preserved. \item Points on the left side of a circline map to points on the left side of the image. \item Angles between intersecting circlines are preserved. \end{itemize} \textbf{Definition: } The point $z^*$ \textit{symmetric} to a point $z$ relative to circle $\bigodot (c,r)$ of center $c$ and radius $r$ is defined to be $ z^* = c + r^2 \frac{z - c}{|z-c|^2}$ . Relative to a line $\ell$, $z^*$ is the reflection of $z$ across the mirror $\ell$. Note that in both cases, the set of fixed points $z^*=z$, are the points on the circline. When $z\ne z^*$ and relative to a line $\ell$, the mirror is the perpendicualr bisector of the segment $[z^* z]$. In the case of a circle $\bigodot(c,r)$, The points $c, z^*, z$ are collinear, and the radius is the geometric mean of their distance form the origin, $r =\sqrt{|z^* -c||z-c|$. Our chief tool for proving some of these properties of MTs is an equation of a circline which works equally well for circles and lines. \section{The Equation of a Circline} For a circle as the points equidistant from its center, we have the equation \[ r^2 = |z-c|^2 = (z-c)\bar{(z-c)} = (z-c)(\bar{z}-\bar{c}) \] which becomes \[ 0 = z\bar{z} - \bar{c}z - c\bar{z} + (c\bar{c} - r^2) = 0.\] For the equation of a line in Cartesian coordinates $x,y$, 3 reals $a,b,d$ with \[ 0 = ax + by +d = (a,b)\bullet (x,y) +d = \Re (a+ib)\bar{(x+iy)} +d .\] Recall that $2\Re(c\bar{z}) = c\bar{z}+\bar{c}z $. We combine the two forms of equations into a single one as follows. \textbf{Theorem:} A circline has the equation, with $m$ complex and $s,t$ real \[ sz\bar{z} + m\bar{z} + \bar{m}z + t = 0, \, |m|^2 > st .\] Note that it would be nicer to define $s$ to be either 0 or 1, but that would make for messier algebra in our application. FILECARD Calculate the invertible relations between the parameters $m,s,t$, and $c,r$ respectively $a,b,d$ in the three equations. \subsection{How MTs Preserve Circlines} Recall that every MT is a composition of Euclidean similarities $z \mapsto az + b $ and possibly the reciprocal function $z \mapsto \frac{1}{z}$. Since preserving geometrical structures is a property that is true of a composition of transformations if it is so for its components, we consider separate cases. \subsubsection{For Taking Reciprocals.} Suppose we substitute $\frac{1}{z}$ for $z$ in the equation of a circline. \begin{eqnarray*} 0 &=& s \frac{1}{z}\frac{1}{\bar{z}} +\bar{m}\frac{1}{z}+ m\frac{1}{\bar{z}} + t \\ 0 &=& s + \bar{m}\bar{z} + m z + tz\bar{z} \\ \end{eqnarray*} Where the second equation was obtained from the first by multiplying through by $|z|^2 = z\bar{z}$. We exclude the case $z=0$ and $z=\infty$, which are reciprocals of each other for other reasons. Thus, the equation we obtain is that of another circline because $|\bar{m}|^2 > ts $ still holds. Moreover $s,t$ switch roles, as do $m,\bar{m}$. Moreover both equations are either true or false together. We conclude that $\frac{1}{z} \in \bigodot(m,s,t)$ if and only if $z \in \bigodot(\bar{m}, t, s)$. \subsubsection{For Euclidean Similarities} Euclidean similarities are compositions of translations, rotations, and dilations, each of which maps circles into circle and lines into lines. So it's not necessary to check these, but it's a good exercise to try it for a translation and dilated rotation separately. FILECARD What is the equation of the translation $z\mapsto z+b$ of a $circline(m,s,t)$? FILECARD What is the equation of the dilated rotation $z \mapsto az$ of a $circline(m,s,t)$? FILECARD Show that under the reciprocal MT, $z\mapsto \frac{1}{z}$, a straight line goes to a circle passing through the origin. Hint: A straight line is a circline which passes through $\infty$. FILECARD What is the reciprocal of a line tangent to the unit circle? \section{MTs Preserve Symmetric Points} Recall an earlier lesson in which we discussed the difference between the customary uses of the term \textit{inversion} both for $z \mapsto z^*$ relative to a circline, and $z\mapsto \frac{1}{z}$. We have been referring to the latter using ``reciprocal" instead. In the case the unit circle as the mirror, the two ideas are closely, and therefore confusingly related. We showed in an earlier lesson that the circlines perpendicular to the unit circle are precisely those with $s=t$. Specializing an the earlier exercise, \[\bigodot(m,1,1)=\bigodot(c,r), \, c=-m, \, r = \sqrt{(|m|-1)(|m|+1)}.\] Hence its reciprocal is the conjugate of itself. In other words, circles perpendicular to the unit circle are mapped to themselves under an involution. \textbf{Theorem:} Let $z^*, z$ be symmetric pairs relative to a circline $ CL = \bigodot(m, s, t)$ and $w=f(z)$ be a Moebius transformation. Then the point $f(z^*)$ is symmetric to $f(z)$ relative to the circline $f(CL)$. We will use the following lemma. \textbf{Lemma:} $CR(z^*,z_0, z_1 , z_2) =\bar{ CR(z,z_0, z_1 , z_2) }$ I have been unable to find a proof that handles circlines together. So we prove this theorem circles and lines separately. Circles come first. \subsection{Proof of the Lemma} Let $\bigodot(c,r) = \bigodot(z_0,z_1,z_2)$. We will tranform the arguments of the CR on the RHS by a composition of elementary MT. \begin{eqnarray*} \bar{ CR(z,z_0, z_1 , z_2)} &=& CR(\bar{z}, \ldots ) \mbox{ ... means do the same to the } z_i \\ \mbox{Applying a translation } &=& CR(\bar{z} - \bar{c}, \ldots ) \\ \mbox{Taking the reciprocal } &=& CR(\frac{1}{\bar{z} - \bar{c}}, \ldots ) \\ \mbox{Usual algebraic trick } &=& CR(\frac{z-c}{|\bar{z} - \bar{c}|^2}, \ldots ) \\ \mbox{Applying a similarity } &=& CR(c + r^2 \frac{z-c}{|\bar{z} - \bar{c}|^2}, \ldots ) \\ \mbox{Apply this MT pts on circle } c+ r^2\frac{z_i-c}{|\bar{z_i} - \bar{c}|^2} &=& c+ r^2\frac{z_i-c}{r^2} = z_i \\ CR(z^*,z_0, z_1 , z_2) &=& CR(c+ r^2\frac{z_i-c}{|z_i - c|^2},z_0,z_1,z_2) \\ \end{eqnarray*} The last line recognizes the fact that MTs (expressed as a crossratio) are one-to-one. \subsection{Proof of the Theorem} Suppose now that we apply an arbitrary MT to each of the 8 entries in the equation we proved in the lemma. The equation remains true or false depending on whether it was true or false before because crossratios are preserved by MTs. But the lemma contains more information. It says that the MT $g(z) = CR(z,z_0,z_1,z_2)$, which sends $\bigodot(z_0,z_1,z_2)$ to the real line, and preserves symmetric points, establishes that $z^* = g^{-1} \bar{g(z)} $. And the points symmetric to the real line (x-axis) are conjugates of each other. So, whenever we can establish the theorem also for points symmetric to a line as follows: Let the mirror line have three points $z_i$ on it, then the MT $g(z)$ sends this line to the real axis, and symmetric paints to symmetric pairs. [to be finished]