Image of the unit circle under inversion

Question

I know that the image of any circle under any Mobius transform is a circle or a line, but I'm having trouble imagining what happens to the unit circle centred at 0 under the mapping $f(z)=1/z$. To me, it looks like you just get everything outside the unit circle.

Answer 1

The unit circle is $\{e^{i t}, \, t\in [0, 2 \pi] \}$. Under $f(z) = 1/z$ it goes to $\{e^{-i t}, \, t\in [0, 2 \pi] \}$, i.e., itself.