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Take for example $f(z)=e^z$, so the inverse is $z(f) = \ln(f) + n\pi i$ for an arbitrarily chosen (but fixed) branch $n\in\mathbb N$. Now if $f$ is restricted to e.g. $0<|1-f|<1$ such that the essential singularity at $f=0$ is not part of the definition region, is $z(f)$ then analytical in that region?

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    yes, you can write down (formally) the power series of the inverse and show it converges or you can write derivatives of $f^{-1}$ in terms of the derivatives of $f$.2011-03-10
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    See http://en.wikipedia.org/wiki/Lagrange_inversion_theorem.2011-03-10
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    @yoyo: you mean yes for the example but in general I have to...2011-03-10
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    ...use the formula linked by @joriki and prove the convergence on a case-to-case basis?2011-03-10
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    The theorem I linked to says "$g$ is analytic at the point $b = f(a)$". That $g$ is analytic implies that its power series converges in some neighbourhood of $b$.2011-03-10
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    @joriki: oh, I missed that line, thanks! so the answer is simply yes, could you post that as an answer so this question doesn't remain unanswered (to the system I mean)?2011-03-10

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See the Lagrange inversion theorem: "$g$ is analytic at the point $b=f(a)$".