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If $h: \mathbb{C} \rightarrow \mathbb{C}$ is a one-to-one holomorphic map, then $h(z)=az+b$, where $a,b\in\mathbb{C}$.

How to prove this argument?

346699
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    What kind of singularity can $h$ have at $\infty$ when $h$ is injective? – Daniel Fischer Dec 05 '14 at 14:37
  • @DanielFischer Thanks for your comment. You can discuss it in different conditions and make this argument more complete. – 346699 Dec 05 '14 at 14:48
  • Duplicate of http://math.stackexchange.com/q/29758/1424, http://math.stackexchange.com/questions/168629/relation-between-linearity-and-injectivity-of-an-entire-function?lq=1, http://math.stackexchange.com/questions/403365/suppose-f-is-entire-and-one-to-one-show-that-fz-azb?lq=1, http://math.stackexchange.com/questions/613575/let-fz-be-a-one-to-one-entire-function-show-that-fz-azb?lq=1 – Jonas Meyer Dec 05 '14 at 15:02
  • WLOG, suppose that $ h(0) = 0 $. Define $ f: \mathbb{C}^{\times} \to \mathbb{C} $ by $ f(z) \stackrel{\text{df}}{=} h ! \left( \dfrac{1}{z} \right) $. Then $ f $ is injective, and $ 0 $ is a singularity of $ f $. By the Weierstrass-Casorati Theorem, $ 0 $ cannot be an essential singularity of $ f $, otherwise $ f[\mathbb{B}(0;r) \setminus { 0 }] \cap f ! \left[ \mathbb{C} \big\backslash \overline{\mathbb{B}}(0;r) \right] \neq \varnothing $, which contradicts the injectivity of $ f $. Hence, the Taylor-series expansion of $ h $ about $ 0 $ has only finitely many non-zero coefficients. – Transcendental Sep 05 '16 at 00:01
  • It follows that $ h $ is an injective polynomial function on $ \mathbb{C} $, so it must be a non-constant linear function on $ \mathbb{C} $. Q.E.D. – Transcendental Sep 05 '16 at 00:04
  • also see this 2013 notes Theorem 5.5 – hbghlyj Aug 22 '23 at 21:51

1 Answers1

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I write $h$ as $f$.

If $f$ is injective, then $f '(P) \ne 0$ for all $P \in \mathbb{C}$ and hence we can define an analytic function $f^{-1}$ from $\text{Im(f)}$ to $\mathbb{C}$. Which means $\mathbb{C}$ and $\text{Im}(f)$ are biholomorphic. If $\text{Im}(f) \ne \mathbb{C}$, then $\text{Im}(f)$ can not be simply connected (By Riemann Mapping Theorem). Hence $\text{Im} (f) = \mathbb{C}$, i.e $f$ is surjective, hence $f$ is a biholomorphism from $\mathbb{C}$ to $\mathbb{C}$. This is in particular a homeomorphism from $\mathbb{C}$ to $\mathbb{C}$, so this must extend to homeomorphism of the one point compactification of $\mathbb{C}$, which implies that $\lim_{z \to \infty} |f(z)| = \infty$. Let $g(z) = h(\frac 1 z)$,so $g$ has a pole at $z= 0$, which means $f$ must be a polynomial, injectivity implies that $f$ must be a linear polynomial.

Shubhodip Mondal
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  • Can you please explain why if Im(f) $\neq \mathbb{C}$ , then why image( f) neednot be simply connected ? –  Sep 01 '20 at 13:00
  • Then Im(f) is biholomorphic to the open unit disk U in C. Thus there exists a invertible holomorphic function g: Im(f)--> U such that gf is a holomorphic function from C ---> U and thus is bounded and therefore gf = constant. This implies the map f: C --> Im f is given by g^-1 (constant) = constant. – Shubhodip Mondal Sep 02 '20 at 02:09