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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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    See also http://math.stackexchange.com/q/29758/2012-07-09

1 Answers 1

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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|. 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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    I am still not convinced about the statement concluded from OMT2013-03-07