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.