Relation between linearity and injectivity of an entire function [duplicate]

Given $f$ entire function on $\mathbb C$ and $f$ one-one. Is it true that $f$ is linear?

At least among polynomials the only such functions are linear!

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marked as duplicate by Prism, amWhy, Mark Bennet, pizza, Jeremy RickardOct 10 '14 at 14:55

Why the hypothesis linear? –  Davide Giraudo Jul 9 '12 at 14:07
this might help you Aneesh math.stackexchange.com/questions/163780/… –  La Belle Noiseuse Jul 9 '12 at 14:28
Pick a point $a$ such that $f\,'(a)\ne 0$. (I don't even want to argue that $f\,'$ never vanishes). Normalize so that $a=0$, $f(0)=0$, and $f\,'(0)=1$. Since $f$ is an open map, there exists $r>0$ such that $\{w:|w|<r\}\subset f(\{z:|z|<1\})$. The function $$g(z)=\frac{f(z)-z}{zf(z)}$$ has a removable singularity at $0$. When $|z|\ge1$, we have $$|g(z)| = \left|\frac{1}{z}-\frac{1}{f(z)}\right|\le \frac{1}{|z|}+\frac{1}{|f(z)|}\le 1+\frac{1}{r}.$$ Thus, $g$ is a bounded entire function. By Liouville's theorem $g$ is constant, and it follows that $f$ is linear.
@JonasMeyer Thanks. After writing this, I noticed that $f$ could be assumed to be merely meromorphic, since the poles of $f$ are removable for $g$. Then the conclusion is $f(z)=z/(1-cz)$, where $c$ is not necessarily zero anymore. –  user31373 Jul 9 '12 at 22:53
$g(z)$ has an infinity plus infinity form so we don't know what it is unless we compute!! $zg(z)$ has the limit 0 as $z$ tends to 0, so the singularity at $0$ is a removable singularity and g extends to an entire function at 0! So don't bother about what $g(0)$ is! The open mapping theorem shows the existence of an $r$ such that $|f(z)|<r$ implies $|z|<1$ Equivalently $|z|\ge1$ implies $|f(z)|\ge r$ or equivalently $|z|\ge1$ implies $\frac1{|f(z)|}\le\frac1r$ At the end of it all, indeed $f(z)=z$ because the absence of poles implies $c=0$ So you are done!! –  Host-website-on-iPage Mar 7 '13 at 4:38