Complex inversion map

Question

How do I show that the map $f: \mathbb{C}\setminus \{0\} \to \mathbb{C}\setminus \{0\} $. $f(z): z \mapsto \frac{1}{z} $ maps circles to either a circle or a line?

Answer 1

Your function f is restriction of a function F from extended complex plane $\bar{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$ to itself, which has this property.

$F:\bar{\mathbb{C}} \rightarrow \bar{\mathbb{C}}$

$F= \frac{az+b}{cz+d}$ where $a=0$, $b=1$, $c=1$, $d=0$

It is common property of Mobius transformation to preserve generalised circles.

There is exact proof of your task on the wiki.

Answer 2

Hint : Try to show that the inverse of $$\{z : |z-a|=r\}$$ is $$\left\{z :\left|z- \frac{\bar{a}}{|a|^2-r^2} \right| = \frac{r}{||a|^2-r^2|} \right\}.$$