The map $\frac1{1-z}$ I want to find the image of the unit circle under the map $\frac1{1-z}$. I first rewrote this as
$$
\frac{1+z}{1-z^2}
$$
But on the unit circle, $z^2 = 1$. So I get $\infty$ for the image of all points on the unit circle as the result. But I thought linear fractional always map circles to lines or circles. Can anyone point out what I did wrong?
EDIT: It turns out that I was confusing $z^2$ with $\bar{z}$, so not all points on the unit circle satisfy $z^2 = 1$. I should let $z = x + iy$ and plug that into the map instead.
 A: The easiest way to find the images of circles and lines (generlized cirlces) under linear fractional transformations (or Möbius transformations) is to simply look at three points on the preimage. Three points on the unit circle is 1, $i$ and -1. Now let
$$f(z) = \frac{1}{1-z}$$
and observe that
$$f(1) = \infty, \quad f(i) = \frac{1+i}{2}, \quad f(-1)=\frac{1}{2}.$$
This shows that the image of the unit circle under $f$ is the straight line through $\frac{1}{2}(1+i)$ and $\frac{1}{2}$, i.e. $\{ z \in \mathbb{C} \, | \, \Re(z) = \frac{1}{2} \}$.
A: Let $w=\frac1{1-z}$. Then, $z=1-\frac1w$ and, with unit circle $|z|=1$
$$|z|^2=(1-\frac1w)(1-\frac1{\bar w})=1\implies w+\bar{w}=1
\implies Re(w)=\frac12$$
which is the vertical line $x=\frac12$ in the complex plane.
A: There are sort of three things going on here:  $1) z\mapsto-z,2)z\mapsto z+1$ and $3)z\mapsto \dfrac1z$.  The first is reflection about the origin, the second translation, and the third inversion.
But personally I prefer to use the method of checking what happens to $3$ points, since generalized circles get sent to generalized circles by Mobius transformations.
