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,
Here's a hack at a proof-sketch.
A bounded entire function is constant -- that's a standard theorem in complex variables.
If your function is unbounded but entire, there must be a sequence $z_i$ with $\lim |f(z_i)| = \infty$, and since the function is bounded on compact sets, we know $\lim |z_i| = \infty$. I'm pretty sure that with a little fiddling, you can conclude that $\lim_{z \to \infty} f(z) = \infty$, so $f$ extends to a function from the Riemann sphere to itself, a function that's 1-1 everywhere (since it sends $\infty$ to $\infty$).
Using this extended function (but still calling it $f$), let $$ g(z) = 1 / f(1/z) $$ and $g$ sends $0$ to $0$, with $g'(0) = a \ne 0$. Then consider
$$ f(z) - \frac{1}{a} z $$
and I'll bet you find that it's a bounded entire function...