Question on holomorphic functions

[disregardthat]

I'm reading through a book on Riemann surfaces, and I tend to get stuck on some of the proofs (maybe because of my lacking background in complex analysis). Anyway, here it goes:

It's a rather lengthy proof, and I'd prefer not to give it in full detail. The question itself is pretty self-contained.

Let h(z) be a holomorphic function from some open subset of $\mathbb{C}$, with h(0) non-zero.

Then why can one find an open disc around 0 such that is $z \mapsto zh(z)$ is biholomorphic onto its image? I.e., why does it have an inverse near 0?

The theorem is about the local behavour of holomorphic functions, basically that one can biholomorphically transform any non-constant holomorphic function to $z \mapsto z^k$ for some integer k > 0 locally around a point.

 

[mathwonk]

this is a corollary of the inverse function theorem. the function g(z) = z.h(z). has derivative at z=0 equal to h(0) ≠ 0, hence looks locally near zero like z-->z.

 

[disregardthat]

That was easier than I thought, thanks mathwonk.

 

[mathwonk]

i recommend cartan's book on holomorphic functions of one and several variables.

 

[disregardthat]

I have Raghavan Narasimhan & Yves Nievergelt Complex analysis in one variable, what do you think of it? Considering getting a different book, maybe Cartans.