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This question already has an answer here:

My instinct is that it may be proved by Liouville's extended theorem. But how to do so? Or are there any other methods?

Thanks!

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marked as duplicate by Julien, Ayman Hourieh, lhf, Potato, Ittay Weiss May 27 '13 at 0:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Consider the image of the unit circle. It has a winding number of 1. $\endgroup$ – Calvin Lin May 26 '13 at 23:59
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    $\begingroup$ If $f(z)$ is not $b + az$ then $f(z) = b + az + cz^2 + \cdots$. And $z\mapsto z^2$ is not one-to-one. $\endgroup$ – Michael Hardy May 27 '13 at 0:02
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    $\begingroup$ @MichaelHardy I don't see how your hint leads to an easy solution. Could you expand? $\endgroup$ – Julien May 27 '13 at 0:11
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    $\begingroup$ @CalvinLin I don't see how that leads to a solution. Could you explain? $\endgroup$ – Potato May 27 '13 at 0:13
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    $\begingroup$ @Potato That's what I call team work. $\endgroup$ – Julien May 27 '13 at 0:14
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Hint: Consider $f(\frac{1}{z})$ and examine the behavior of the singularity at $0$. It can't be an essential singularity (see this using the Weierstrass-Casorati theorem on the behavior near an essential singularity along with the open mapping theorem). So it is either removable or a pole. Then $f(\frac{1}{z})$ has a finite number of negative powers of $z^{-1}$ in its Laurent expansion, so $f(z)$ is a finite power series and must be a polynomial. If the polynomial has degree greater than $1$, then it has $2$ or more roots, contradicting the one-to-one hypothesis.

Note that this gives you the automorphism group of $\mathbb C$.

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  • $\begingroup$ There are 2 or more roots...but what if the roots are in the same location, so that we have one distinct root with multiplicity = n? Doesn't this keep f as a one-to-one function, @Potato? $\endgroup$ – User001 Dec 18 '14 at 5:58
  • $\begingroup$ @LebronJames If it has a root with multiplicity $2$, then locally at that root it looks like $w=z^2$, which is clearly not injective. Ahlfors provides a rigorous explanation of this somewhere his book. Essentially, WLOG we may assume the root is at zero, so $f(z)=z^2(1+\dots)$, and then you can take a square root locally since the second factor is nonzero around $0$. $\endgroup$ – Potato Dec 18 '14 at 6:47
  • $\begingroup$ Awesome stuff, @Potato. I saw this in the summer - what a timely reminder from you :) Thanks so much. Have a great night. $\endgroup$ – User001 Dec 18 '14 at 7:13
  • $\begingroup$ @LebronJames My pleasure. Good luck with your studies. $\endgroup$ – Potato Dec 18 '14 at 11:00

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