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I'm introducing myself to Complex analysis and Möbius transformations and I read that Möbius transformations map circles and lines to circles and lines.

Are there any other functions that are not Möbius transformations but they can map circles to circles?

If I know that $f(z)$ maps a circle to another circle, can I assume that $f(z)$ is a Möbius transformation?

mzdravkov
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To elaborate on sometempname's comment: if $f(z)=\overline{z}$, then for the circle $|z-a|=r$, we have $$|f(z)-\overline{a}|=r,$$ so the image of a circle is a circle. Similarly, the image of a line is a line, so this will have the desired property but not be a Mobius transform.

Samir Khan
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I presume you talk about analytic maps. But even then you may take products of Möbius transformations which also maps $S^1=\{|z|=1\}$ to itself (1-1). Such transformations are called: Blaschke products

If you do not require 1-1 then you also have maps like $z\mapsto z^p$ and if you require analyticity only in a neighborhood of $S^1$ there are many more.

On the other hand a map that always maps any circle or line to a circle or a line is either a Möbius transformation (whence meromorphic) or a Möbius transformation composed with complex conjugation. Perhaps this is more what you are after... (and a proof is not that difficult)

H. H. Rugh
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  • $z^2$ maps the circle $|z-1|=1$ to a cardioid. – sometempname Aug 21 '16 at 20:28
  • True, I only considered the last question and thought of the standard unit circle. – H. H. Rugh Aug 21 '16 at 20:33
  • Is there a reference for that last result? I have frequently seen the forward direction “Mobius transforms map generalized circles to generalized circles” but I have almost never seen “the only continuous maps that send generalized circles to generalized circles are mobius transforms” or perhaps the more constructive “if a map sends generalized circles to generalized circles then we derive that it must be a Mobius transformation” – Sidharth Ghoshal May 31 '24 at 03:10
  • I remember having seen it somewhere, but can't find it now. As I recall you better assume some smoothness, like C^1 or else it gets messy. For a C^1 map it goes something like: (1) Show that the differential map at a non-singular point must preserve circles (2) Conclude that the map is a rational map of the sphere (3) Show that the map can have no critical points (at a point where the derivative vanishes, the map can not preserve circles) (4) Conclude that the map is Möbius. Without doubt there are other ways. – H. H. Rugh Jun 02 '24 at 08:51