I am working with the function $f(z) = \frac{z+1}{z-1}$, for a complex variable $z$. I understood that for $z$ in the unit disc, i.e $\lvert z\rvert \le 1$, $\mathrm{Re}(f(z)) \le 0$.


What if $z$ is in a disc, say $D = \left\{ z \in \mathbb C: \, \lvert z\rvert \le a \right\}$ for a given constant $a \in \mathbb R\_{+}$?


Note that the function is its own inverse, if it matters …. I'm not very good with analysis and I have trouble visualizing functions from $\mathbb C$ to $\mathbb C$, since there are 4 dimensions ….



Image of a complex disc by this function?
This question might have been better suited to math.SE but since it's here I may as well post an outline of an answer. Your function $f$ is an example of a Moebius transformation, and these are known to take circles to circles provided that you allow straight lines as circles passing through "the point at infinity". So for $0<a<1$ the image of your closed disc is another closed disc, living in the left half-plane; for $a>1$ you get the complement of some open disc that lives in the right half-plane.


