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Well, I was trying to prove $Aut(\mathbb{C})=\{f\in H(\mathbb{C}): f(z)=az+b\}$ for some constant $a,b$, As $f$ is entire these three possibilities can happen

$1$. $f(z)=a_o$(constant)

$2$. $f(z)=\Sigma_{n=0}^{k}a_nz^n$ ($a_k\neq 0,k\ge 1)$

$3$. $f(z)=\Sigma_{n\ge 0} a_nz^n$, $a_n\neq 0$ for infinitely many $n$

$1$ is not possible as $f$ is injective and $2$ is just a polynomial of degree $k$ by Fundamental Theorem of Algebra $0$ has $k$ preimage in $\mathbb{C}$ so this is also not possible. I would like if some one tell me about $3$ in detail with explanation, also do I need to consider about transcendental entire functions? I have no clear Idea about them. Please help.

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Option 3 is transcendental functions. –  Cameron Buie Jun 27 '12 at 14:54
    
It is a corollary of Picard Theorem. –  y zhao Jun 27 '12 at 14:57
    
Look at $f(1/z)$ and see what you can say about the singularity at the origin. Or see the answers to this question. –  Per Manne Jun 27 '12 at 15:23
    
could u tell me why $f(1/z)$ has an essential singularity at $0$ if $f(z)$ is not a polynomial? why it should have singularity at all? –  Une Femme Douce Jun 27 '12 at 15:45
    
@Mex it's essentially a definition of an essential singularity. If you ask why it can't behave reasonably (e.g., be bounded), it's because then $0$ would be a removable singularity, and therefore $f(1/z)$ would have a Taylor expansion $\sum_{n=0}^\infty a_n z^n$. So it would have two different Laurant expansions in $0<|z|<r$, which is impossible (the Laurant coefficients are uniquely determined by the function). –  user31373 Jun 27 '12 at 16:53

1 Answer 1

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You need to know a few things about how holomorphic functions behave near isolated singularities here.

Let $g(z)=f(1/z)$, then $g$ is a holomorphic function except at the point $z=0$ where it is not defined. By definition, this is a singularity for $g$.

If the singularity is removable, then $g$ is bounded near the origin, so $f$ is bounded outside some large disk. It follows that $f$ is a bounded entire function, and by Liouville's theorem $f$ is a constant.

If $g$ has a pole of order 1 at the origin then $g(z)=\sum_{k=-1}^{\infty}a_kz^k$, so $f(z)=\sum_{k=-1}^{\infty}{a_k\over z^k}$. Since $f$ is regular at $z=0$ we must have $a_k=0$ for $k\geq 1$, so $f(z)=a_0+a_{-1}z$ is a linear function.

If $g$ has a pole of order $k>1$ at the origin, then a similar argument as above gives that $f$ is a polynomial of order $k>1$, which contradicts that $f$ is injective, as you have pointed out already.

If $g$ has an essential singularity at the origin, then there are different arguments you can use. The shortest is Picard's big theorem, which says that $g$ will take all possible values infinitely many times, with at most one exception, on any punctured neighborhood of the origin. This will obviously contradict the requirement that $f$ is supposed to be injective.

A weaker result on $g$, which can also be used, and which is much simpler to prove than Picard's theorem, is the Casorati-Weierstrass theorem. It says that if $g$ has an essential singularity at the origin then the image of any punctured neighborhood of the origin will be dense in $\rm\bf C$. This will again be impossible to reconcile with the requirement that $f$ is supposed to be injective.

To prove the Casorati-Weierstrass theorem, assume that $U$ is a punctured neighborhood of $0$ and that $b$ is such that $|g(z)-b|>\epsilon$ for all $z\in U$. Then $h(z)={1\over g(z)-b}$ will be bounded on $U$, so it will have a removable singularity at $z=0$. If this leads to a value $h(0)\neq 0$ then $z=0$ will be removable for $g(z)$, and if we get $h(0)=0$ then $|g(z)|\to\infty$ as $z\to 0$, so $z=0$ is a pole for $g(z)$. In both cases this contradicts that $z=0$ is supposed to be an essential singularity for $g$. Hence $g(z)$ has to come arbitrarily close to $b$ on $U$.

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