TITLE: Let $f(z)$ be a one-to-one entire function, Show that $f(z)=az+b$.
QUESTION [5 upvotes]: Let $f(z)$ be a one-to-one entire function, Show that $f(z)=az+b$.
My try :
Because $f$ is entire it has a taylor series around zero (in particular).
$f(z)=\sum^{\infty}_{k=0} a_kz^k$
Proof by contradiction :
let $m \geq 2$
Suppose $a_m \neq 0 $ and $ f(c)=f(b) \Rightarrow \ \ \  0= f(c)-f(b)= \sum^{\infty}_{k=0} a_k(c^k-b^k ) \therefore \ \ c^m-b^m=(c-b)(c^{m-1}+c^{m-2}b+...+b^{m-2}c+b^{m-1})=0 $ 
This implies :
i) $c=b \ \ \ $ seemingly I have not got a contradiction
or
ii)  $(c^{m-1}+c^{m-2}b+...+b^{m-2}c+b^{m-1})=0$ 
Can somebody tell me what is going wrong ?
Thanks in advance

REPLY [3 votes]: First we claim that $f(z)$ has to be polynomial. If not, then in its Taylor expansion there are infinitely many non-zero terms, and hence $f(1/z)$ has an essential singularity at $0$. By Big Picard, in a neighbourhood of zero, we hit every complex value (except possibly one) infinitely many times, contradicting injectivity of $f$. It follows $f(z)$ has to be a polynomial.
Can you take it from here?