# Entire one-to-one functions are linear

Can we prove that every entire one-to-one function is linear?

You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.
To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard’s theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f(\{z:|z|>n\})$ is dense in $\mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire’s theorem, $D=\bigcap_n f(\{z:|z|>n\})$ is dense in $\mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.
I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f(\{z:|z|<1\})$ is open, and $f(\{z:|z|>1\})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.