I'm reading Silverman's Complex variable with application.

at page 78, the author says "We say that $s$ and $s^*$ are the inverse points with respect to circle in $\mathbb{C}$ if every line or circle passing through $s$ and $s^*$ intersect at right angles."

and the author consider the case of unit circle and a point $z$ lying inside the unit citcle. And he claimed that the inverse point of $z$ is $\frac{1}{\bar{z}}$. He show this as follows:

Let $L$ be a line passing through the center $O$ (the origin) and $z$. Draw a line $S$ perpendicular to $L$ through $z$. This line intersects unit circle twice. Draw tangent lines at which $S$ intersects the circle. Then the two tangent line intersect at the line $L$. The point on the $L$ at which two tangent line intersect is $\frac{1}{\bar{z}}$. So $O$ and $z$ and $\frac{1}{\bar{z}}$ are collinear.

My questions are like this:

  1. What did he mean by "every line or circle passing through two inverse points intersect at right angles"? What are the objects which intersect each other? If you want to say "intersect", there has to be at least two objects. Does it mean that given circle and any arbitrary circle or line passing through two inverse poinrs intersect orthogonally? I can't imagine how it might look like.

  2. and The phrase "Every line passing through two inverse points" looks weird at least to me, since if we have two points on the plane, the line passing through those points is uniquely determined. Why did he talk like that?

  3. Why if you follow the above construction then come out $\frac{1}{\bar{z}}$? I don't get it. Actually I'd like to see the algebraical construction corresponding above geometrical construction. (e.g. starting from setting z and arrive at $\frac{1}{\bar{z}}$)

  • 2
    $\begingroup$ He means that every line/circle through $s$ and $s^*$ meet $C$ at right angles. $\endgroup$ – Lord Shark the Unknown Apr 22 at 7:02
  • $\begingroup$ If you like, you can consider a line to be a circle of infinite radius and say, "Every circle passing through $s$ and $s*$ ... " though personally, I find this less clear than the original statement. $\endgroup$ – saulspatz Apr 22 at 7:49
  • $\begingroup$ @LordSharktheUnknown What do you refer "$C$"? the circle in $\mathbb{C}$ to which the two points are inversed(?)? $\endgroup$ – Grimza Apr 22 at 8:16

It might be easier to start with the converse statement. If a circle $C'$ through $z$ and $z^*$ intersects the unit circle $C$ at $z_1$, then $|z_1|^2 = 1 = |z| \, |z^*|$, therefore the radius of $C$ at $z_1$ is a tangent to $C'$ by the tangent-secant theorem. The construction with the line $S$ follows because if $S$ intersects $C$ at $z_1$, the segment $[z_1, z^*]$ is a diameter for the circle through $z, z^*, z_1$.

"intersect at right angles" is supposed to be "intersects the unit circle at right angles" and "tangents at which $S$ intersects the unit circle" is supposed to be "tangents at the points at which $S$ intersects the unit circle". And $|z - z_0| \, |z - z^*| = R^2$ should be $|z - z_0| \, |z^* - z_0| = R^2$.

  • $\begingroup$ Thank you. I eventually understood it. And there are a lot of typo... $\endgroup$ – Grimza Apr 22 at 23:10

Perhaps a picture will help. I have just eyeballed these two points, but it appears that circles for which both points lie on the circle intersect the black circle at right angles.

enter image description here


The problem you are bothered with (question 3 in your post): let there be $z$, find such a number $z'$, that $|z| \cdot |z'|$ = 1. i.e. it lies on a unit circle. and $|z| = \alpha |z'| $ for real $\alpha$ i.e. they lie on the same line. This point is unique, so let's prove that $z' = \frac{1}{\bar{z}}$.

$|\frac{1}{z}| = \frac{1}{|z|}$ but it has an angle that is opposite to the needed, so we need to take the complex conjugate, and we are done. Let's prove that actually the task I formulated is your case.

Lemma: $\cos(\theta) = |\frac{r_1^2 + r_2^2 - d^2}{2 r_1r_2}|$ where $\theta$ is an angle of intersection, of two circles, $r_1$ and $r_2$ are radii and $d$ is the distence between centers.

$r_1 = 1$ $ r_2 = \frac{z + \frac{1}{\bar{z}}}{2} $ try to calculate the angle yourself. That should work.

That was a calculation of a circle that has a diameter ending with $z$ and $z'$. but you can construct a parametric family of circles that come through those two points, using determinant method and assigning the coordinates of the 3rd points as parameters. There should be a shorter way to do it.

This is a picture from Аргунов Б.И., Балк М.Б. Геометрические построения на плоскости. Пособие для студентов педагогических вузов. This is a geometric way to construct an inverse point, to which I am used to.

Not an answer, but some thoughts. You probably heard about Mobius transformation . This is a generalization of inverse transform. They are conformal (i.e angle preserving) maps. See for some basic literature to discover more. If we take all coefficients to be real, then we get a map from $PGL(2, \mathbb{R}) \to \text{automorphisms of upper-half plane} $. $z \mapsto f(\bar{z})$.

Let us coincide Mobius maps of kind $z \mapsto \frac{az+b}{cz+d}$ with matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. This matrix set is called Linear Group $GL(2 , \mathbb{C})$ If you multiply two fractions, the coinciding matrix of the result is the product of two. Note, that matrix $\begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}$ is the same as $\begin{pmatrix}5 & 0 \\ 5 & 5 \end{pmatrix}$ So, to make the map one-to-one, we need to look on the matrices with determinant $ad - bc = \pm 1$. This gives us Projective Linear Group $PGL(2,\mathbb{C})$.

Let's try to look at the matrices of determinant one. It's Special linear group $SL(2, \mathbb{C})$There are two matrices $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $-I = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ that give equal maps, so we take $PSL(2, \mathbb{C})= SL(2, \mathbb{C})/\pm I$

I see that it is a useful information not really needed here.

one of ways to construct an inverse point.

  • $\begingroup$ Thank you for your answer. What is $PSL(2,\mathbb{R})$? $\endgroup$ – Grimza Apr 22 at 23:11

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