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

**Answer**

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 C for each positive integer n. By the open mapping theorem, the set is open. By Baire’s theorem, D=⋂nf({z:|z|>n}) is dense in 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.

**Attribution***Source : Link , Question Author : Petey , Answer Author : Michael Hardy*