Complex Analysis: Inverse function is holomorphic

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Homework Statement


The problem is from Sarason, page 44, Exercise IV.14.1.

Let f be a univalent holomorphic function in the open connected set G, and let g be the inverse function.
Assume that f(G) is open, that g is continuous, and that f\prime\neq 0\forall z\in G. Prove g is holomorphic.

Homework Equations


The Attempt at a Solution


I've tried 3 different methods, failing with each. I'd appreciate any insight into the problem.
My professor recommended looking at the calculus proof of the derivative of an inverse, and the following:
Start with w_{0}\in f(G), and with h small, look at \frac{(g(w_{0}+h)-g(w_{0}))}{h}.

Because f(G) is open, we know there is a \delta>0\ni \left|h\right|<\delta gives us w_{0}+h\in f(G).
So with h that small, we have w_{0}+h=f(z) for some z \in<br /> G.
 
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You're off to a good start. I'd suggest also defining z_0\equiv f^{-1}(w_0) and then using that along with the other defined variables to write out the definitions, in terms of \epsilons and \deltas, of f being differentiable, and f' and g' being continuous.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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