# Every bijective conformal transformation of $\mathbb{C} \to \mathbb{C}$ is of the form $f(z)=az+b$. [duplicate]

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?

• 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. Apr 29, 2019 at 9:08
• For your last question see en.wikipedia.org/wiki/Conformal_map Apr 29, 2019 at 9:09
• I warn you that analyticity does not make the question in the title very easy. It is quite non-trivial Apr 29, 2019 at 9:10
• @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 Apr 29, 2019 at 9:10
• @kneidell so are you asking what I find funny? I thought by definition, conformal is biholomorphic Apr 29, 2019 at 9:11

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.
• it says in the problem that $f$ is bijective Apr 29, 2019 at 9:21
• also, i don't know what $U$ is Apr 29, 2019 at 9:21
• @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. Apr 29, 2019 at 9:23
• @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. Apr 29, 2019 at 9:46