Conformal circles (A property of inversions)

Question

I came across the following proposition in this book:

If C is a circle in the Euclidean plane, i_C is conformal, that is it preserves angles. Also, i_C takes circles not containing the center of C to circles, circles containing the center to lines, lines not containing the center to circles containing the center, and lines containing the center to themselves.

I could not figure out what this proposition means. I need to understand this statement before I move on to its proof.

Answer 1

A holomorphic mappingg $\mathbb{\hat{C}} \rightarrow \mathbb{\hat{C}}$ that has a holomorphic inverse is called a conformal mapping of $\mathbb{\hat{C}}$. It turns out that every conformal mapping of $\mathbb{\hat{C}}$ is a Möbius transformation.

The proposition you ask about asserts that a conformal map will send the points along a circle to either a circle or a line (depending on whether or not $0$ is on the circle) and sends the points on a line to lines or circle (depending on whether or not $0$ is on the circle). I give a relatively easy to follow proof of this below.

Theorem Any Möbius transformation is a composition of translations, dilation and inversions. (A Möbius transformation is a map of the form $\phi(z) = \displaystyle\frac{az+b}{cz+d}$.) A translation is of the form $\phi(z) = z+c$, a dilation is of the form $\phi(z) = \lambda z$, and an inversion is of the form $\displaystyle \phi(z) = \frac{1}{z}$.

Proof Let $\phi$ be given. If $\phi(\infty)=\infty$, then we know that $\phi(z) = a z+b=(z→z+b)◦(z→a z)$. Suppose $\phi(∞)=c \neq ∞$. Then $\psi=(z→1/z)◦(z→z−c)◦φ$ is a Möbius transformation fixing $\infty$: $$\infty \xrightarrow{\psi} c \xrightarrow{z-c} 0 \xrightarrow{1/z} \infty$$ Thus, $ψ$ is a composition of translations, dilations and inversions. The same is true for $φ$.

Now that there is some background on the conformal maps we move to the actual problem at hand.

Theorem Möbius transformations preserve the class of circles and lines.

Proof Circles and lines are clearly preserved by translations and dilations. It suffices to check that they are preserved under inversion. Consider a circle $C=\{z∈\mathbb{C}||z−a|=r\},\ r>0$. Put $w=1/z$. \begin{align*} z∈C&⇔|z−a|^2=r^2\\ &⇔|z|^2−\overline{a} z−a\overline{z}+|a|^2=r^2 \end{align*} If|a|=r (equivalently, C goes through $0$), then \begin{align*} z∈C&⇔|z|^2−\overline{a} z−a\overline{z}=0\\ &⇔1−\overline{a w}−a w=0\ \text{or}\ w=∞\\ &⇔2Re(a w)=1\ \text{or}\ w=∞\\ &⇔Re(e^{iarg(a)}w)=\frac{1}{2|a|}\ \text{or}\ w=∞ \end{align*} This last condition describes a line.

If $|a|\neq r$(equivalently, C does not go through 0), then \begin{align*} z∈C&⇔|z|^2−\overline{a} z−a\overline{z}+|a|^2−r^2=0\\ &⇔1−\overline{a w}−a w+(|a|^2−r^2)|w|^2=0\\ &⇔|w|^2−\frac{\overline{a w}+a w}{|a|^2−r^2}+\frac{1}{|a|^2−r^2}=0\\ &⇔|w|^2−\frac{\overline{a w}+aw}{|a|^2−r^2}+\frac{|a|^2}{(|a|^2−r^2)^2}+\frac{1}{|a|^2−r^2}−\frac{|a|^2}{(|a|^2−r^2)^2}=0\\ &⇔\left|w-\frac{|a|^2}{|a|^2−r^2}\right|^2−\frac{r^2}{(|a|^2−r^2)^2}=0 \end{align*} The last equation describes a circle. A similar calculation shows that under inversion, the image of a line $\ell$ is a circle (if $0∉ \ell$) or a line (if $0∈\ell$).

Answer 2

Let the centre of $C$ be $y$, and the radius of $C$ be $s > 0$. Inversions in $C$ preserve the single unioned set of lines and circles. That is, every circle is mapped to a line or a circle, and every line is mapped to a line or a circle. As to which it maps to depends on where $y$ lies relative to the mapped circle/line.

Let $D$ be a circle centred at $x$, of radius $r$. If the centre of $C$ (i.e. $y$) does not lie on $D$ (i.e. $\|x - y\| \neq r$), then $i_C(D)$ is still a circle. Otherwise, if the centre of $C$ lies on $D$ (i.e. $\|x - y\| = r$), then $i_C(D)$ is a line (actually, it technically should be $i_C(D \setminus \{y\})$, since $y$ is not in the domain of $i_C$, but this is usually brushed over).

On the other hand, let's look at a line $L$. If $y \notin L$, then $i_C(L)$ is a circle. If $y \in L$, then $i_C(L)$ is a line (and is actually equal to $L$).

The proposition also keeps track of whether $y$ lies in the resulting set or not, but this can be deduced from the given information and the fact that $i_C$ is an involution, meaning $i_C^{-1} = i_C$. So, for example, it's clear that, when $y \notin L$, $i_C(L)$ would have to be a circle containing $y$, because such a circle when mapped again under $i_C$, would have to return to the line $L$ (if it didn't, it would map to a circle, and spoil $i_C$ being an involution).

As for the conformal remark, you can view the angle between two intersecting differentiable curves in terms of the angle of intersection between their tangents (or equivalently, normals). A conformal map like $i_C$ will preserve differentiability (for points other than $y$), and when the image of two such curves meet, the tangents may be different, but the angles between them will be the same. This is what conformal maps do.

(Interestingly, these inversion maps work more generally in higher dimensional spaces, and even infinite-dimensional Hilbert spaces. They still preserve spheres and hyperplanes!)