Is the image of the open disk under a linear-fractional transformation always a Caratheodory Domain?

Question

I am not enough of a complex analyst to understand well the definition of Caratheodory Domain... It seems to me that set with a boundary that looks like a small deformation of a circle or loop would fit the description. If I take a linear-fractional transformation which is a self-map of the open unit disk (things like (x+1)/2, 1/(2-x), or very specifiically, x/(2-x))... are their images of the open disk considered Caratheodory Domains?

Answer 1

The definition of a Caratheodory domain seems rather subtle. However, we have a simple example of a Caratheodory domain, the bounded region bounded by a closed Jordan curve. Now, if $\phi$ is a homeomorphism of the Riemann sphere that does not map any point in the domain to $\infty$, then the image of $\phi$ is again the bounded region bounded by a Jordan curve.

Therefore, if we have a region bounded by a circle $C$, and $c z + d \ne 0$ for all $z$ inside, or on $C$, then the map $z\mapsto \frac{a z + b}{c z + d}$ maps $C$ to a circle, and its inside to the inside of the image ( so again a C domain).