Proof of conformal property for circle inversion

Question

I'm reading a College Geometry: A Unified Development [unfortunately not available through google book preview], and I came across the Theorem, that circle inversions preserve angles between two arbitrary intersecting curves

The proof goes about proving, that the angle $\theta$ between the curves $C_1$ and $C_2$ (depicted in $\color{red}{\text{red}}$ and $\color{blue}{\text{blue}}$ colors respectively in the figure) is the same as the angle $\theta'$ between the images of those curves under the circle inversion.

$t_1$ and $t_2$ (colored lines) are tangents to $C_1$ and $C_2$ curves respectively.

$t_1'$ and $t_2'$ circles are images of $t_1$ and $t_2$ under the inversion (straight lines are mapped to circles through $O$ - the center of circle of inversion).

Dashed lines near the $O$ are tangents to circles, and the solid lines near the second point of circle intersection ($P'$) - are tangents to mapped curves.

The proof goes by claiming that $\varphi$ - the angle between dashed tangent lines to $t_1'$ and $t_2'$, is equal to $\theta$ (one of the properties, that were previously proved, is that tangent line to the circle at $O$ is parallel to its image under the inversion) [so far so good]

... and the angle $\theta'$ between the tangent lines to the $C_1'$ and $C_2'$ mapped curves equals the angle between the circles ($t_1'$ and $t_2'$) at point of intersection, which in turn equals to $\varphi$ ...

So, once I know that black non-dashed lines are indeed tangent to circles I'm done. But why are those lines tangent to circles?

Answer 1

Since the question asks for a clarification of a textbook proof, I'll post copies of the proof and diagram, because the answer will refer to them.

The question states that "the lines near the second point of circle intersection - are tangents to mapped curves." However, the proof in the textbook says that the straight lines at $P'$ are tangents to the circles $t'_1$ and $t'_2$,which are the images of the original curve tangents. The confusion comes, I believe, from the phrase "corresponding tangents $t'_1$ and $t'_2$ at $P'$ to the image curves $C'_1,C'_2$" Here they mean that the circles are tangent to the curves, and they seem to be making an implicit assumption that if curves touch ("are tangent to each other") then the image curves will also touch ("be tangent").

So it remains to show that the tangents to the image circles are the same as the tangents to the image curves at $P'$. This can be done by treating the tangents as the limits of secants going through $P'$, and it is a fairly straightforward exercise to show that the circle secants and curve secants converge to the same straight line.

There is a traditional (and in my opinion canonical) proof involving secants that most other texts use to show the anti-conformal property of inversions. Wolfe's Introduction to Non-Euclidean Geometry, pg. 240 gives one version. In case this link isn't stable, here's a screen cap:

Answer 2

You are referring to two circles passing through the center of inversion which have straight line inversions by complex Inversion transformation

$$ z_2= \frac {a^2}{z_1}=\frac {1}{z_1} $$

The sketch shows Inversion of Circles passing through Inversion center O. The green circle is the Circle of Inversion. I is image of point P passing through O, i.e., points P on blue circle map to points I on red circle across mirror points M of the green inverting mirror circle.

Shown are parts of angle

$$ \alpha = \alpha_1+ \alpha_2 $$

at the three intersection /concurrent points $(A_1,A_2,O)$ and how they should be same at the opposite corner. (I have not drawn tangents to avoid crowding of lines and circles meeting at corner).

From a differential equation viewpoint, $(\sin \psi= r/c) $ for a circle through center. With inversion transformation $ r \rightarrow \dfrac{a^2}{r}$ it becomes $ ( r \sin \psi = a^2/c) $ which is a straight line of shortest length or geodesic.

Shall revise my answer if it is not addressing your question properly. Blue circles cut the inversion circle and second picture is zoomed.

Half sum of the three angles at center is $180^{\circ}$ which is also seen as sum of three internal angles of a triangle. This demonstrates angle sum of the vertices.

Hope you are seeing a bigger/comprehensive picture here.

Answer 3

It is not difficult to use multi-variable calculus to show that an inversion map is conformal (e.g., showing that the Jacobian matrix at each point of an inversion map is a scalar multiple of a reflection matrix). Let's say that you have an inversion $\iota$ about the point $(h,k)$ with radius $r$. Then $$\iota(x,y)=\left(h+\frac{r^2(x-h)}{(x-h)^2+(y-k)^2},k+\frac{r^2(y-k)}{(x-h)^2+(y-k)^2}\right).$$ The Jacobian matrix of $\iota$ is then $$J(x,y)=\frac{r^2}{(x-h)^2+(y-k)^2}R(x,y)$$ where $R(x,y)$ is the reflection matrix $$R(x,y)=\begin{bmatrix}-\cos\theta(x,y) &-\sin\theta(x,y)\\ -\sin\theta(x,y)&\cos\theta(x,y)\end{bmatrix}$$ s.t. $$\cos\theta(x,y)=\frac{(x-h)^2-(y-k)^2}{(x-h)^2+(y-k)^2}$$ and $$\sin\theta(x,y)=\frac{2(x-h)(y-k)}{(x-h)^2+(y-k)^2}.$$ Indeed, if a differentiable bijection $\phi:M\to M$ where $M$ is an open subset of $\Bbb R^n$, is such that the Jacobian matrix of $\phi$ at each point $s\in M$ is of the form $\sigma(s)A(s)$, where $\sigma:M\to\Bbb R$ is a strictly positive-valued function and $A(s)$ is an orthogonal matrix for each $s\in M$, then $\phi$ is a conformal map.

Here is a differential geometry argument. I merely intended to give the thread some answer. I do not immediately see an elementary geometry proof, since the discussion of tangents of an arbitrary curve should involve some level of calculus/differential geometry. Please enlighten me if there is a way to avoid calculus/differential geometry completely.

An inversion map $\iota$ can be extended to the Riemann sphere $\hat{\Bbb C}=\Bbb C\cup \{\infty\}$. (Say, $\iota$ is the inversion about a circle centered at some point $o$. Then $\iota(\infty)=o$ and $\iota(o)=\infty$. A straight line is actually a circle in $\hat{\Bbb C}$ that passes through $\infty$, and the inversion w.r.t. this straight line is the same as the reflection about the line.) This map can be easily seen to be smooth, but this is the pivotal part of the problem so an interested reader should check it. Since $\iota$ is an involution, it is a auto-diffeomorphism of $\hat{\Bbb C}$.

Let $I$ be the open interval $(-1,1)$. Suppose that the curves $\alpha,\beta$ which are, respectively, the images of two differentiable embeddings $a,b:I\to\hat{\Bbb C}$ are tangent at a point $p=a(0)=b(0)$, then the tangent spaces $T_p\alpha=da_0(T_0I)$ and $T_p\beta=db_0(T_0I)$ coincide. (Differentiability of $a$ and $b$ is necessary. Otherwise it makes no sense to discuss the term tangents.)

Let $q=\iota(p)$. Then the differential map $d\iota_p:T_p\hat{\Bbb C}\to T_q\hat{\Bbb C}$ is an isomorphism of vector spaces due to $\iota$ being a diffeomorphism of $\hat{\Bbb C}$. Therefore $d\iota_p$ sends $$T_p\alpha\mapsto T_q\iota(\alpha)$$ and $$T_p\beta\mapsto T_q\iota(\beta).$$ But $T_p\alpha=T_p\beta$, so $$T_q\iota(\alpha)=T_q\iota(\beta).$$ Hence the inversion images $\iota(\alpha)$ and $\iota(\beta)$ of $\alpha$ and $\beta$ under $\iota$ meet at $q$, at which the tangent spaces coincide. Thus $\iota(\alpha)$ and $\iota(\beta)$ are tangent at $q$.

Back to the problem, when $\alpha$ is the original blue curve and $\beta$ is the original green line, then $\alpha$ and $\beta$ are tangent at $p$. Then the blue image curve $\iota(\alpha)$ and the green image circle $\iota(\beta)$ are tangent at $q=\iota(p)$. Therefore the black line that is tangent to the green circle $\iota(\beta)$ at $q$ is also tangent to the blue image curve $\iota(\alpha)$. The same applies if $\alpha$ is the original red curve and $\beta$ is the original violet line.

From the previous paragraph, it follows that if $\iota$ is conformal with respect to straight lines and circles in $\hat{\Bbb C}$ (well in fact straight lines in $\hat{\Bbb C}$ are the same as circles), then $\iota$ is conformal with respect to all differentiable curves in $\hat{\Bbb C}$. Since conformality w.r.t. straight lines and circles is known (and not difficult to establish via elementary geometry), we are done.

It is possible to establish conformality of $\iota$ solely using differential geometry as well. Let $g$ be the usual Riemannian metric on $\hat{\Bbb C}$. Show that the pullback $\iota^*g$ of $g$ under $\iota$ satisfies $(\iota^*g)_p=f(p)g_p$ for some function $f:\hat{\Bbb C}\to \Bbb R$ s.t. $f(p)>0$ for every $p\in\hat{\Bbb C}$.