# Let $f(z)$ be a one-to-one entire function, Show that $f(z)=az+b$.

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).

$f(z)=\sum^{\infty}_{k=0} a_kz^k$

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 ?

#### Solutions Collecting From Web of "Let $f(z)$ be a one-to-one entire function, Show that $f(z)=az+b$."
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.
It does not follow from $\sum_1^\infty a_k(c^k-b^k)=0$ that $c^m-b^m=0$ for all $m$. This is why your proof attempt does not work.