Conformal map of the complex plane is linear

Asked Dec 27 '12 at 03:22

Active Dec 27 '12 at 03:43

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Gameline Complex Analysis, P. 265 #8 is like this,

Show that every conformal self-map of the complex plane $ \mathbb C$ is linear.

Hint: The isolated singularity of $f(z)$ must be the simple pole.

First of all, how do I argue that the singularity is not essential or removable? Second of all, how do I argue it is a pole and is simple?

I can see there is a singularity at $ \infty$ because function is not really defined there. Some hints please!

Addendum Now I can see that the singularity can not be removable because of the Liouville's Theorem. If the singularity at $\infty$ were removable then that would make function bounded and hence analyticity in the entire complex plane tells function is constant but which is impossible because function is bijective.

edited Dec 27 '12 at 03:43

asked Dec 27 '12 at 03:22

Deepak

1,616

The key point is that $f(z)$ is bijective, so if it had a pole of order $2$ or an essential singularity at infinity can you see how injectivity would fail? – JSchlather Dec 27 '12 at 03:27
Not really, would you please elaborate? – Deepak Dec 27 '12 at 03:29
If it has a pole of order 2, then the the principle part of the Laurent series has two negative terms, does that imply that $z_{0}$ maps to $\infty $ twice or what? Confused!! – Deepak Dec 27 '12 at 03:33
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The following are links to related questions: one two three four five – Jonas Meyer Dec 27 '12 at 05:23
Thanks Jonah Meyer. That really helps. – Deepak Dec 27 '12 at 06:17

Conformal map of the complex plane is linear

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