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This question already has an answer here:

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, Namaste, Mark Bennet, user147263, Jeremy Rickard Oct 10 '14 at 14:55

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The link given by @Patience leads to a proof, but one can avoid the heavier things like Picard, Casorati-Weierstrass and the very notion of essential singularity. Liouville's theorem is enough.

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.

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    $\begingroup$ This is a great answer. Of the answers to this question on other threads which I have seen, this is most similar to this answer of Zarrax. $\endgroup$ – Jonas Meyer Jul 9 '12 at 22:47
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    $\begingroup$ @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. $\endgroup$ – user31373 Jul 9 '12 at 22:53
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    $\begingroup$ @Patience You have a function such that $f\,'(a)\ne 0$. Let $\tilde f(z)=(f(z+a)-f(a))/f\,'(a)$. This new function has $\tilde f(0)=0$, and $\tilde f\,'(0)=1$, as desired. Note that $f$ is linear if and only if $\tilde f$ is linear. So, we can simply work with $\tilde f$ from now on, and even rename it as $f$ to simplify writing. $\endgroup$ – user31373 Jul 27 '12 at 1:28
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    $\begingroup$ $f(z)=\frac{z}{1+cz}$ is linear? $\endgroup$ – Marso Mar 6 '13 at 23:16
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    $\begingroup$ $g(z)$ 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!! $\endgroup$ – Karthik C Mar 7 '13 at 4:38

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