Can it be shown that analytic functions map disks to disks?

Question

I have read that analytic functions are special in that they 'map disks to disks', as unlike, say, smooth functions, the mapping they represent is a composition of rotation and scaling. So given an analytic function $f$ and a disk D with center $(a, b)$ and radius $R$, that is $$D = \{(x, y) \in \mathbb{R}^2: (x-a)^2 + (y-b)^2 < R^2\},$$

how can we prove that the mapping of the disk under the analytic function $f$ must be of the form $f(D) = \{(x, y) \in \mathbb{R}^2: (x-c)^2 + (y-d)^2 < S^2\}$?

Answer 1

The stated claim is false. For example, here is the image of the disk $|z| \leq 1/2$ under the map $z \mapsto z^2 + z + 1$:

Answer 2

The Riemann mapping theorem says: if $D$ is a simply connected region in $ \mathbb C$ with $D \ne \mathbb C$, then there is a function $f:D \to \mathbb D= \{z \in \mathbb C: |z|<1\}$ such that:

1. f is analytic

and

2. f is bijectiv.

Cosequence: $f^{-1}(\mathbb D)$ is in general not a disc.