On inversive geometry

Question

I am given the following problem set:

Observe the circle $K$ with center $0$ and radius $r$ in the complex plane $ \mathbb{C} \simeq \mathbb{R}^2$. Show that the inversion on $K$ is given by the following mapping: $$ i_K : \mathbb{C} \rightarrow \mathbb{C} \\ z \mapsto \frac{r^2}{\bar z}$$

First thoughts : I know that inversions have to fulfill the following properties

• $\forall p \in E\smallsetminus \lbrace z \rbrace: i_K(p)\in \overrightarrow{zp} \ \ $ where $z$ is the origin of circle $K$

• $d(z,p) \cdot d(z, i_k(p)) =r^2$

Problem : Since we just introduced the inversion (with no introduction to hyperbolic geometry) I don't know how to prove the two properties - which I think are necessary to prove the mapping. I am thankful for any kind of help and advice.

Answer 1

Solution :

We have to fulfill the following two properties that the mapping $$ i_K : \mathbb{C} \rightarrow \mathbb{C} \\ z \mapsto \frac{r^2}{\bar z}$$ states an inversion on a circle $K$

(a) $\forall p \in E\smallsetminus \lbrace z \rbrace: i_K(p)\in \overrightarrow{zp} \ \ $ where $z$ is the origin of circle $K$

(b) $d(z,p) \cdot d(z, i_k(p)) =r^2$

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(a) $$i_K(p) = \frac {r^2}{\bar p} = \underbrace{\frac{r^2}{\vert p \vert^2}}_{\text{scalar}} \cdot p $$ so this shows us that $i_K$ lies on $\overrightarrow{zp}$

(b) \begin{align}d(z,p) \cdot d(z, \frac {r^2}{\bar p}) &= r^2 \\ d(z,p) \cdot d(z, \frac{r^2}{\vert p \vert^2} \cdot p) &= r^2 \\ \vert p\vert \cdot \left \vert \frac{r^2}{\vert p \vert^2} \cdot p\right \vert = r^2 \cdot \left \vert \frac{\vert p \vert^2}{\vert p \vert^2} \right \vert &= r^2 \end{align}

and we are done.

Answer 2

Recall that $z\bar{z} = |z|^2$. It may be helpful to observe that

$$ \bar{z} = \frac{|z|^2}{z} $$

Hyperbolic geometry, incidentally doesn't have anything to do with the inversive plane.

Answer 3

You just need to write out what both properties look like with this transformation.

The first one says "$\frac{r^2}{\overline{p}}$ is an $\Bbb R$ scalar multiple of $p$". This is true since $\frac{r^2}{\overline{p}}=p\frac{r^2}{p\overline{p}}$.

The second says $|p||\frac{r^2}{\overline{p}}|=r^2$, which you can easily confirm using the properties of the modulus on complex numbers.