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How can it be shown that $$Aut(\mathbb{C})=\{f|f(z)=az+b,a\neq 0\},$$ where an automorphism of $\mathbb{C}$ is defined as a bijective entire function with entire inverse?

If $f$ is of the form $f(z)=az+b$, with $a\neq 0$, then obviously $f$ is in $Aut(\mathbb{C})$. How can I prove the converse?

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It depends on how much complex analysis you know. Let me give you a short proof (using heavy machinery).

If $f$ has an essential singularity at $\infty$, it can't be injective on $\mathbb{C}$ due to big Picard. If the singularity at $\infty$ is removable, then $f$ is constant by Liouville. Hence $f$ has a pole at $\infty$ which means that $f$ is a polynomial. If $\deg f \ge 2$, it won't be injective (fundamental theorem of algebra). That only leaves you with polynomials of degree exactly 1.

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    $\begingroup$ f(z)=z^3 has deg=3 but is injective $\endgroup$ Feb 19 '13 at 20:52
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    $\begingroup$ @FedericaMaggioni No, $f(z) = z^3$ is not injective: $f(1) = f(e^{2\pi i/3}) = f(e^{-2\pi i/3}) = 1$. $\endgroup$
    – mrf
    Feb 19 '13 at 20:59
  • $\begingroup$ It is worth noting, perhaps, that the fundamental theorem itself doesn't suffice if there is a single root of maximal multiplicity. You need to utilize something like Hurwitz's theorem to account for these cases. $\endgroup$
    – Bar Alon
    Dec 4 '18 at 12:01
  • $\begingroup$ Single root of maximal multiplicity means that the polynomial is of the form $a(z-z_0)^n$, and one can show directly that this is not injective. $\endgroup$
    – mz71
    Feb 2 '19 at 11:28
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    $\begingroup$ One can apply fundamental theorem to $f’$ and conclude $f$ cannot be injective $\endgroup$ Nov 26 '20 at 4:25
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If $f : \mathbb{C} \rightarrow \mathbb{C}$ is an automorphism, it is entire and hence has a globally convergence power series $f(z) = \sum_{n=0}^\infty a_n z^n$. We can extend this function to a map $\tilde{f} : \mathbb{C} \rightarrow \bar{\mathbb{C}}$ by considering it as a function into the extended complex plane $\bar{\mathbb{C}} = \mathbb{C} \cup \lbrace \infty \rbrace$ which has a singularity at $\infty$.

If we take a coordinate chart about $\infty$ as $U_\infty = \lbrace \infty \rbrace \cup \lbrace z : |z| > 1 \rbrace$ with coordinate map $\phi : U \rightarrow \mathbb{D}$ given by $\phi(z) = 1/z$ for $z \neq \infty$ and $\phi(\infty) = 0$ then we can explore the nature of $\tilde{f}$'s singularity at infinity by composing it with $\phi$:

$ \tilde{f}(\phi(z)) = \sum_{n=0}^\infty \frac{a_n}{z^n} $

Its almost by definition that the above formula has an essential singularity at $z=0$ unless the $a_n$ eventually are all zero. If there were an essential singularity then by Cassorati-Weirestrass $f$ maps $U_\infty \setminus \lbrace \infty \rbrace$ to a dense subset of $\mathbb{C} $. But then $ f(\mathbb{D}) $ would necessarily intersect that dense set non-trivially, so $f$ would not be injective and hence not an automorphism. Therefore the above series $\tilde{f}(\phi(z))$ is a rational function, which implies that $\tilde{f}$ is a polynomial, so $f$ is a polynomial.

Finally, by the fundamental theorem of algebra, a polynomial is injective if and only if its degree is one, so $f(z) = az + b$.

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You can start by noting that Big Picard's Theorem will tell you that you can not have an essential singularity at infinity because otherwise you will find multiple points mapping to the same value in different neighborhoods of infinity. This tells you that $f$ is necessarily a polynomial

Then note that if a function is degree greater greater than 1 it will have multiple roots and therefore have more than 1 value mapping to 0. Then you just eliminate constant functions to keep injectivity.

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There is a nice way to do this by using Riemann surfaces, plus the fact that one can view $\Bbb C\mathrm{P}^1$ as "$\Bbb C$ with an added point at infinity" or equivalently, we may view $\Bbb C$ as $\Bbb C\mathrm{P}^1\setminus \{\infty\}$. The key observation is that, if one has an automorphism $f$ of $X\setminus P$, where $P=\{p_1,\dots,p_n\}$ is a finite set and $X$ a Riemann surface, then it must extend to an automorphism of $X$ (not necessarily unique). This is done by taking little disk charts around a puncture $p_j$ and holomorphically continuing $f$ over the point, which is possible by boundedness.

Now, it is a standard result in the theory of Riemann surfaces that the meromorphic function on $\Bbb C\mathrm{P}^1$ are the rational function (i.e. quotients of polynomials), and that meromorphic functions can be viewed as holomorphic maps into $\Bbb C\mathrm{P}^1$. Hence, one just needs to check which rational functions are actually biholomorphisms. Because of the fact that a non-constant holomorphic map (between Riemann surfaces) is locally always of the form $z^n$ for some $n\geq 1$, it suffices to check injectivity. Using this, one establishes easily that the automorphisms of $\Bbb C\mathrm{P}^1$ are precisely the Möbius transformations $$ f(z)=\frac{az+b}{cz+d} \qquad \qquad ad-bc\neq 0$$ where $z\in \Bbb C\mathrm{P}^1=\Bbb C\cup \{\infty\}$ (so the domain (and range) include the point "$\infty$"). Now, the above shows that any automorphism of $\Bbb C$ extends to a Möbius transformation which must send $\infty$ to itself. This is equivalent to $c=0$, and you obtain the result you're looking for.

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    $\begingroup$ Can you explain your boundedness comment (on why you can continue any automorphism of $\mathbb{C}$ to $\mathbb{CP}^1$)? Thank you! $\endgroup$
    – user062295
    Nov 2 '17 at 2:50

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