Conformal map of the complex plane is linear

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.

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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 3:27
Not really, would you please elaborate? –  Deepak Dec 27 '12 at 3:29
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 3:33
The following are links to related questions: one two three four five –  Jonas Meyer Dec 27 '12 at 5:23
Thanks Jonah Meyer. That really helps. –  Deepak Dec 27 '12 at 6:17