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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 $f(D) = \{(x, y) \in \mathbb{R}^2: (x-c)^2 + (y-d)^2 < S^2\}$?

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    All analytic functions map disks to disks.? – 2017-02-09
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    Well that's just completely false. – 2017-02-09
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    I would guess that to the contraire it is quite unusual that analytic function maps disks to disks in general. I wonder why you came up with that hypothesis? – 2017-02-09

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

enter image description here

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    It says on this page that they map disks to disks - https://www.johndcook.com/blog/2013/08/20/why-are-differentiable-complex-functions-infinitely-differentiable/ – 2017-02-13
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    @csss That quote is being imprecise and, taken literally, is wrong. It should say this: as the diameter of a disk in the domain of the function shrinks to $0$, the image of that disk in the range of an analytic function resembles a disk more and more (but is not necessarily actually a disk at any point). One fact that can be proven is that the only meromorphic functions $f$ which map every circle to another circle have the form $f(z) = (az+b)/(cz+d)$. – 2017-02-13
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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

  1. f is bijectiv.

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