circlescomplex-analysisgeometrymobius-transformation
I'm trying to prove that the inversion mapping $f(z) = \frac{1}{z}$ sends circles or lines to circles or lines.
Apparently the set $$\{z \in \mathbb{C}: |z-a|^2 = r^2 \}$$ describes either a circle or a line.
How does this set includes circle and lines?
Circles are defined as $|z – a| = r$ and lines are defined as $|z-a| = |z-b|$?
One possible method is to find points $a, b$ which are symmetric with respect to both the given circle $C$ and the given line $l$, and then choose $T$ as a Möbius transformation which sends $a, b$ to $0, \infty$: $$ T(z) = \frac{z-a}{z-b} $$ Möbius transformations preserve symmetry, therefore $0, \infty$ are symmetric with respect to both $T(C)$ and $T(l)$, which are therefore circles centered at zero.
The symmetry of the problem suggest to search for real $a, b \in \Bbb R$. Then symmetry with respect to the line $\operatorname{Re} z = 5$ means $a + b = 10$, and symmetry with respect to the circle $\vert z \vert = 4$ means $ab = 4^2$.
So $a, b$ are the solutions of the quadratic equation $x^2 - 10 x + 16 = 0$, which can easily be determined.
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:
Best Answer
That set describes a circle, not a line.
The function $f$ maps circles which pass through the origin into lines and all other circles into circles. It als maps line passing through the origin into lines and all other lines into circlse.
Consider, for instance, the line $\{t+(1-t)i\,|\,t\in\mathbb R\}$. For each $t\in\mathbb R$, you have$$\frac1{t+(1-t)i}=\frac{t+(t-1)i}{t^2+(1-t)^2}=\frac t{2t^2-2t+1}+\frac{t-1}{2t^2-2t+1}i$$and$$(\forall t\in\mathbb R):\left(\frac t{2t^2-2t+1}-\frac12\right)^2+\left(\frac{t-1}{2t^2-2t+1}+\frac12\right)^2=1.$$