This question already has an answer here:
- Entire one-to-one functions are linear 6 answers
Show that if f is entire and one-to-one, then it must be of the form AZ+B, with A not equal to zero.
I am editing my question, since there are duplicates on this forum to the question of why an entire and one-to-one function must be of the form AZ+B, with A non-zero.
I am currently stuck at f(z)=AZ for A non-zero, from using Liouville's Theorem on g=z/f(z).
I'd like to show that f(z) cannot also take the form AZ^2, AZ^3, and so on...
I think that is done by using the Fundamental Theorem of Algebra and saying that an nth degree polynomial in z (with non-zero coefficient, A) has exactly n roots. But I'm thinking about the situation when all of the roots are at one location, so that we have only one distinct root with multiplicity = n. Then this doesn't rule out the case that f(z) is one-to-one.
What can I do to show the polynomial must only be of degree 1? (I've seen some derivative arguments now, including @JohnHuges ' argument below, where f' is not zero, but I don't understand this argument and why we can conclude from this that f is not one-to-one...)
Thanks in advance,