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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?

Bunbury
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    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. – kneidell Apr 29 '19 at 09:08
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    For your last question see https://en.wikipedia.org/wiki/Conformal_map – Kavi Rama Murthy Apr 29 '19 at 09:09
  • I warn you that analyticity does not make the question in the title very easy. It is quite non-trivial – Kavi Rama Murthy Apr 29 '19 at 09:10
  • @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 – kneidell Apr 29 '19 at 09:10
  • @kneidell so are you asking what I find funny? I thought by definition, conformal is biholomorphic – mathworker21 Apr 29 '19 at 09:11
  • @mathworker21 your'e citing the conformal mapping theorem.. this is not the definition of a conformal mapping – kneidell Apr 29 '19 at 09:14
  • @kneidell dude I speak English. I am saying I thought that was the definition. and I just checked; I learned complex analysis from the book of Stein and Shakarchi, and they define a conformal map as a biholomorphism. and what are you talking about w.r.t. the conformal mapping theorem??? the conformal mapping theorem says you can map any simply connected domain biholomorphically into the disk. how would that say that conformal mappings are biholomorphic? I respectfully think you have no idea what you're talking about. – mathworker21 Apr 29 '19 at 09:19

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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.

Kavi Rama Murthy
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  • it says in the problem that $f$ is bijective – mathworker21 Apr 29 '19 at 09:21
  • also, i don't know what $U$ is – mathworker21 Apr 29 '19 at 09:21
  • Thank you, Kavi! This is very helpful! – Bunbury Apr 29 '19 at 09:23
  • @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. – Kavi Rama Murthy Apr 29 '19 at 09:23
  • @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. – mathworker21 Apr 29 '19 at 09:46