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In complex analysis textbooks $\mathrm{Aut}\,(\mathbb C)$ frequently serves as an example of conformal automorphism groups, which is said to consist of all linear functions $az+b$ with $a\neq0$. Equivalently it means that any biholomorphism $\mathbb C\to\mathbb C$ is of the form $az+b$ where $a\neq0$.

However, all textbooks that I have say this fact is obvious but I cannot see it, so I want to ask how to prove it?

josephz
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    I would be very surprised if an introductory complex analysis textbook actually claimed this was obvious, without providing more context (e.g., a result that it is a corollary of). – Eric Wofsey Sep 24 '17 at 13:43
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    https://math.stackexchange.com/questions/307185/automorphisms-of-the-complex-plane – Martin R Sep 24 '17 at 13:45
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    I wouldn't say it's obvious. Usually you look at the nature of the singularity at $\infty$; see https://math.stackexchange.com/questions/307185/automorphisms-of-the-complex-plane. The easiest proof is big Picard, which tells you about the behavior of the function at $\infty$ – Marcus M Sep 24 '17 at 13:45

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