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Show that every bijective conformal transformation of $\mathbb{C} \to \mathbb{C}$ is of the form $f(z)=az+b$.

I find it difficult to start because it does not specify $f$ to be analytic, which would have made the question a lot easier.

Is it true that bijective conformal mappings are necessarily analytic with nonzero derivatives?

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    $\begingroup$ Conformal means that locally, the map looks like the composition of a homothety and a rotation of $\mathbb C$. Write this condition down in terms of the partial derivatives of $f$ (considered as a map from $\mathbb R^2$ to itself) to see that this implies the Cauchy-Riemann equations. $\endgroup$
    – kneidell
    Commented Apr 29, 2019 at 9:08
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    $\begingroup$ For your last question see en.wikipedia.org/wiki/Conformal_map $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 29, 2019 at 9:09
  • $\begingroup$ I warn you that analyticity does not make the question in the title very easy. It is quite non-trivial $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 29, 2019 at 9:10
  • $\begingroup$ @mathworker21 I honestly don't see what's so funny here.. the connection between conformal and analytic is not trivial, especially not to a student.. I think this is a very fine question $\endgroup$
    – kneidell
    Commented Apr 29, 2019 at 9:10
  • $\begingroup$ @kneidell so are you asking what I find funny? I thought by definition, conformal is biholomorphic $\endgroup$
    – mathworker21
    Commented Apr 29, 2019 at 9:11

1 Answer 1

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Hints (for this non-trivial result): I am assuming that $f$ is analytic. Please see my comment above. I will only use injectivity of $f$. The range of $f$ is simply connected. If it is not $% %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion $ it would be conformally equivalent to the open unit disk $U$ and we get a contradiction by invoking Liouville's Theorem. Thus $f$ is onto $% %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion $. Prove (using Picard 's Theorem) that $f(\frac{1}{z})$ cannot have an essential singularity at $0$. It cannot have a removable singularity either so it has a pole. This forces $% f$ to be a polynomial and it must of degree one since it is one-to-one.

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  • $\begingroup$ it says in the problem that $f$ is bijective $\endgroup$
    – mathworker21
    Commented Apr 29, 2019 at 9:21
  • $\begingroup$ also, i don't know what $U$ is $\endgroup$
    – mathworker21
    Commented Apr 29, 2019 at 9:21
  • $\begingroup$ Thank you, Kavi! This is very helpful! $\endgroup$
    – Bunbury
    Commented Apr 29, 2019 at 9:23
  • $\begingroup$ @mathworker21 Added definition of $U$. Assuming that $f$ is bijective instead of just injective doesn't make it much simpler. You still need Picard's Theorem. $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 29, 2019 at 9:23
  • $\begingroup$ @KaviRamaMurthy 1). the question says bijective, so people might be confused reading your answer. 2). maybe I'm missing something, but why introduce the letter $U$? You only use it once. 3). how did you deduce that you need Picard's theorem. did you secretly prove that the problem statement easily implies Picard's theorem? If not, seems a bit egotistical to say that your solution is the only one. $\endgroup$
    – mathworker21
    Commented Apr 29, 2019 at 9:46

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