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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

the8thone
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    This seems like a really hard way to prove it. Try considering the type of $\infty$ as a singularity of the function $f$. –  Dec 20 '13 at 03:09
  • @T.Bongers If $\infty$ be a removable singularity of $f(z)$ it means that $\lim_{z \rightarrow \infty} f(z) = w \neq \infty$ and by liouville's theorem f(z) is a constant and we're done in this case (right?) – the8thone Dec 20 '13 at 03:19
  • Yes, if the singularity is removable, $f$ is constant. So the other two options are a) pole, and b) essential singularity. –  Dec 20 '13 at 03:20
  • @T.Bongers If $\infty$ be an essential singularity, i.e. $\lim_{z \rightarrow \infty} f(z)$ does not exist do we get a contradiction to the fact that $f$ is entire ? – the8thone Dec 20 '13 at 03:26
  • @T.Bongers Should I consider the principal part of the Laurant series at $\infty$ ? That's so strange to me ... – the8thone Dec 20 '13 at 03:27
  • @T.Bongers In the first place "What" caused you think about the type of singularity of $\infty$ , don't tell me "experience" !!! – the8thone Dec 20 '13 at 03:29
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    I've seen this proof before a few times; one way to see why it might be useful is to know that the sort of singularity does a lot to characterize the function: It's true in general that if a function has a pole at $\infty$, then it's a polynomial. If you can prove this, you're almost done. –  Dec 20 '13 at 03:31
  • @T.Bongers I assumed that $f(z)$ has a pole of order $k$ at infinity, i.e. $f(\frac{1}{z})$ has a pole of order $k$ at zero, i.e. $f(\frac{1}{z})=\frac{g(z)}{z^k}$, where $g(0) \neq 0$ and $g(z)$ is analytic in asufficiently small nbhd of $0$, i.e I can write its taylor series $$f(\frac{1}{z})=\frac{g(z)}{z^k}=\frac{a_0}{z^k}+\frac{a_1}{z^{k-1}}+...+ a_k + a_{k+1}z +...$$ if I somehow show that $a_n=0$ for all $n>k$ I'm done with proving what you said. But how can you claim that ? can you give me another hint ? – the8thone Dec 20 '13 at 04:15
  • Also see http://math.stackexchange.com/questions/29758/entire-1-1-function – Ayman Hourieh Dec 20 '13 at 22:08

2 Answers2

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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?

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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.

Potato
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