Local homeomorphism question

1. Feb 4, 2009

bdeln

I've just been reading about how complex functions can be defined on the extended complex plane. They start with rational functions as examples, and defining them at oo so they're continuous at oo in a sense. Eg, 1/z would be defined to be 0 at z = oo.

I understand that given a holomorphic function f, then f is not a local homeomorphism whenever f'(z) = 0 (open mapping theorem right?), but I'm wondering, what if f'(z) = oo now? Or what if I get a situation where f'(z) isn't even defined at oo, which seems to happen a lot? For example, f(z) = z^2 isn't differentiable at oo if I understand this stuff correctly, but is differentiable everywhere else. I get the feeling that, since oo is a fixed point of z^2, and z^2 is going to be covering everything twice near oo, it's not going to be a local homeomorphism there either .. but I'm wondering if anyone could clear this up for me properly, or point me in the direction of some notes/books that deal with this stuff nicely.

Thanks.