The Image of Hyperbola Under Inversion

Question

Consider a hyperbola given by the equation $x^2-y^2=1$. I want to find its image under the inversion $f(z)=\frac{1}{z}$. Can I use this as the following argument?

First, I can observe if $z=re^{i\theta}$, then $f(z)=\frac{1}{r}e^{-i\theta}$. Then, I can observe that $x^2-y^2=1$ can be written in terms of polar coordinates as $r^2\cos(2\theta)=1$. So, as every point $(r,\theta)$ maps to $(\frac{1}{r},-\theta)$, then the equation $r^2\cos(2\theta)=1$ will become $$(\frac{1}{r})^2\cos(2(-\theta))=1$$ which gives us $$r^2=cos(2\theta)$$ i.e. the graph of lemniscate.

Answer 1

Let $w=u+iv=\frac{1}{z}$, $z=x+iy \implies u+iv=\frac{\bar z}{z \bar z} \implies u=x/(x^2+y^2), v=-y/(x^2+y^2)~~~~(1)$ $$x^2-y^2=1=Re[z^2]=Re[\frac{1}{w^2}]=1 \implies Re[\frac{\bar w^2}{(\bar w w)^2}]=\frac{u^2-v^2}{(u^2+v^2)^2}=1~~~~(2)$$

So the image of $x^2-y^2=1$ under inversion $(w=1/z)$ is the curve $u^2-v^2=(u^2+v^2)^2$, which can be finally written as $$x^2-y^2=(x^2+y^2)^2.$$

Answer 2

Your proposition is quite correct. In fact it belongs to the set of Mclaurin? curves.

$$ r^n= a ^n \cos( n\cdot \theta ) $$ Obtained by exponentiation of $ z = a e^{i \theta}$ to power of $n$ and then taking real part.

The subsets are also the inversions got by $$ r\to \dfrac{a^2}{r}$$

$ n=2, -2 $ are the Lemniscate and equi- hyperbola $ x^2-y^2= a^2$

$ n=1, -1 $ are the straight lines and circles through origin

$$r= a \sec \theta,r= a\cos \theta $$

The above and all the Rosettes formed this way may be termed Mclaurin Rosettes. Iirc it was mentioned in the second chapter among exercise examples in Heinrich Guggenheimer's of Differential Geometry text book... with reference to both aspects: Inversions and their generation from real part of exponentiated unit complex number.