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?

  • $\begingroup$ What if $f$ is not continuous? I'm sure you can find some counterexample. $\endgroup$
    – user223391
    Aug 21 '16 at 20:17
  • 3
    $\begingroup$ $f(z)=\overline{z}$ $\endgroup$ Aug 21 '16 at 20:18
  • $\begingroup$ @sometempname That's an answer. =) $\endgroup$
    – Pedro Tamaroff
    Aug 21 '16 at 20:19
  • $\begingroup$ Do you require that every circle maps to a circle, or just that there exists one circle that is mapped to a circle? $\endgroup$
    – mrf
    Aug 22 '16 at 7:21
  • $\begingroup$ @mrf the latter one $\endgroup$
    – mzdravkov
    Aug 22 '16 at 18:42

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.


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)

  • $\begingroup$ $z^2$ maps the circle $|z-1|=1$ to a cardioid. $\endgroup$ Aug 21 '16 at 20:28
  • $\begingroup$ True, I only considered the last question and thought of the standard unit circle. $\endgroup$
    – H. H. Rugh
    Aug 21 '16 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.