Lesson on Symmetric Points
Z4 1may11, 14jul11\\ \textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{What is needed before this lesson: inversion, symmetric point} Caution! \br This lesson is still under construction. I refers to Hvidsten's Section Geometric Properties of Moebius transformations. \section{Introduction} In this lesson we study the Symmetry Principle [Hvidsten Proposition 8.9] to discover the relation of \textit{inversions in circles} to Moebius Transformations. This lesson treats three important concepts: \subsection{MTs preserve circlines.} This means that MT takes every circle and every line into another circle or line. The proof gives us an opportunity of writing the equation of lines and circles in the plane which differs from what you learned (and taught, perhaps) in high school. \subsection{Symmetry relative to circlines} The second concept is that of symmetry across a circle, which generalizes the concept of symmetry across a line. Symmetry across a line is also called \textit{ reflection in the line}, and you know from high school analytic geometry, at least for the line $y=x$ where $(x,y) \mapsto (y,x)$. This approach permits us to "cover" the theory of inversions in circles in an efficient way, using the MTs and CRs instead of the more cumbersome but less demanding approach using nothing but analytic geometry. \subsection{ MTs are conformal } Finally, we look briefly at a geometric property of an MT, namely that it preserves the measure and orientation of angles. This section is, in a sense, totally an "asside". It is the only place in the course where we actually need t the calculus. And, we do so for efficiency, as usual. This property can be proved purely with analytic geometry, using trig identities, like the Law of Cosines. But this approach comes at a much greater effort and tedious calculations. In general, a transformation of the plane (or the Riemann sphere) \textit{ conformal} if it preserves angles. We will show that MTs are conformal \textbf{because} they have a non-zero complex derivative. But since that is the only property we will use, there are other transformations which are conformal, but not MTs. We will not pursue this further. It belongs into a proper course on complex variables. \section{Meditation on just what an "equations of $\Xi$" really is.} You may take this paragraph as a bit of remediation, especially if you have forgotten what me mean by "the equation of $\Xi$" where $\Xi$ is some geometric object. The phrase itself, "the equation of $\Xi$" is shorthand for \textbf{the equation(s) the coordinates of a point need to satisfy in order for the point to be in the set} $\Xi$. Aren't you glad you don't have to say this all the time. \subsection{Equation of a circle} Let's take the equation of a circle with center at $(p,q)$ and radius $r$. Using Euclid's definition of a circle as \textit{ the locus of points in the plane equidistant from its center} we write \[ circ(p,q,r) =\{(x,y) | (x-p)^2 + (y-q)^2 = r^2 \} . \] (In your Journal, underline the part of this expression actually is the "equation of the circle".) By multiplying out the equation we obtain a quadratic equation of the form \[x^2 + y^2 + ax + by + c = 0.\]Question 1.Find the $a,b,c$ in terms of the $p,q,r$.Question 2.Given $a,b,c$, how do you find $p,q,r$. What conditions must $a,b,c$ fulfill for there to be a solution? \section{Equations of Circlines} Now imagine what calculations you would have to go through to determine whether four given points line on the same circle? Recall that in high school geometry you learned how to determine the center and radius of the circle through 3 given points.Question 3.Describe how to construct the center and radius of the circle through $P,Q,R$. The solution in the Complex plane is surprisingly simple. \textbf{ Theorem } The equation of a circline is given by \[ \rho z\bar{z} + \mu \bar{z} + \bar{\mu}z + \sigma =0, \rho \ge 0, z\bar{z} \gt \rho\sigma.\] \subsection{What you can see immediately from the equation} \subsubsection{Comparing two complex numbers.} Note that the inequality $\rho \ge 0$ also says that $\rho$ is a real number, i.e. $\mathfrak{Im}(\rho)=0$. And, once again, remember that some of the complex numbers $z$ are \textit{real}, i.e. $\mathfrak{Im}(z)=0$, some are \textit{pure imaginary}, i.e. $\mathfrak{Re}(z)=0$, but most of them are neither real nor imaginary, i.e. they don't lie on either axis. In general, there is no way to define the comparison $z > w$ for complex numbers, unless they both are real. (We could, but don't compare them if they are both pure imaginary.) \subsubsection{Detecting additional real numbers} That $\sigma$ must also be real, given that $\rho$ is real, can be deduced from the equation itself as follows. Note that $z\bar{z} = |z|^2$ is real. And, because $\bar{\mu}z$ and $\mu \bar{z}$ are conjugates, their sum is twice their real parts, which is real. So, even though $z$ and $\mu$ are arbitrary complex numbers, when you solve the equation for $\sigma$, it equals a real number, and so it itself real.Question 4.Finish the argument that if $\rho \ge 0$ then $\sigma$ is real as well. \subsection{Discovering the circline satisfying the equation.} \subsubsection{When this circline is a straight line.} We easily dispose of the case when $\rho =0$, for then, writing $z=x+iy$ and $\mu=\alpha + i\beta$ and subsituting yields a linear equation.Question 5.Write the linear equation in $x,y,\alpha,\beta,\sigma$ your subsitution resulted in. \subsubsection{When it is this circline is a circle.} For $\rho \gt 0$ we divide through by it, writing $m := \frac{\mu}{\rho}$ and $s:= \frac{\sigma}{\rho}$ for easier calculation. \begin{eqnarray*} z\bar{z} + m \bar{z} + \bar{m}z + m\bar{m} &=& m\bar{m} - s \\ (z + m)(\bar{z} + \bar{m}) &=& m\bar{m} - s \\ |z + m|^2 &=& |m|^2 - s \\ |z - (-m)|^2 &=& r^2 \\ \end{eqnarray*} The last being the equation of a circle centered at $-m$ and radius $r$.Question 6.Show why $|m|^2 - s$ is positive, and hence it has a positive square root $r$. \section{Applications} One immediate appliation of writing the equations of a circline in the symmetric form above is the following elegant proof of the theorem: \textbf{Theorem: The inversion of a circline in the unit circle is another circline.} \subsection{Algebraic forms for the inversion in the unit circle.} Recall that the function which takes \[ z \mapsto z^* := \frac{1}{\bar{z}} = \frac{z}{|z|^2} \] is called the \textit{ inversion in the unit circle}. Note that $z^{**}=z$. Note that $|z|=1 \implies z* = z $. So the inversion leaves every point on the unit circle fixed. The unit circle acts as a mirror. A point and its image under the inversion are said to be \textit{ symmetric with respect to the unit circle }. So, $0$ and $\infty$ are symmetric with respect to the unit circle. \subsection{What happens to the equation under inversion in the unit circle} We now argue as follows. Suppose $z^*$ lies on the circline $\ell^*$ with parameters $\sigma, \mu, \rho$, i.e. we substitute $z^*$ into the equation: \begin{eqnarray*} \sigma z^*\bar{z^*} + \mu \bar{z^*} + \bar{\mu}z^* + \rho &=& 0 \\ \sigma \frac{1}{\bar{z}}\frac{1}{z} + \mu\frac{1}{z} + \bar{\mu}\frac{1}{\bar{z}} + \rho &=& 0 \\ \sigma + \mu \bar{z} + \bar{\mu}z + \rho z \bar{z} &=& 0 \\ \end{eqnarray*} which is exactly the equation of the circline $\ell$ with parameters $\rho, \mu, \sigma$. From this we see that $z$ also satisfies the equation of a circline, but one with $\rho$ and $\sigma$ interchanged. Because we know the equation of a circline and its inversion in the unit circle, and we know the geometry of the circline from its equation, we can deduce many additional properties. \begin{enumerate} \item If $\rho \sigma \ne 0$ then both $\ell$ and $\ell*$ are circles, centered at $- \frac{\bar{\mu}}{\rho}$ and $- \frac{\bar{\mu}}{\sigma}$ respectively. \item Therefore, the centers of these two circles are collinear with the origin. They lie on the ray from $0$ to $-\bar{mu}$. \item If $\rho = 0$, then $\ell$ is a straight line, and $\ell^*$ is a circle that passes through the origin. \end{enumerate}Question 7.Show that if $\rho=\sigma=1$ then $\ell^* = \ell$. Obviously the centers are the same $-\mu$. Compute their respective radii to see that they are the same too. Note that a circline passes through the origin if and only if $z=0$ is a solution of its equation. And this happens exactly when the constant 3rd parameter in its equation is zero.Question 8.Show that if a circline passes through the origin, then it's inverse in the unit circle is a circline that passes through $\infty$ by showing that its equation is that of a straight line. \section{Summarizing} We close the first part of the lesson with a question. Do we have enough information now to see that the case that $\sigma=\rho=1$ are the circlines that are perpendicular to the unit circle?