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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 $f$. – yoyo Mar 10 '11 at 15:36
@yoyo: you mean yes for the example but in general I have to... – Tobias Kienzler Mar 10 '11 at 16:38
...use the formula linked by @joriki and prove the convergence on a case-to-case basis? – Tobias Kienzler Mar 10 '11 at 16:39
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 $b$. – joriki Mar 10 '11 at 16:45
up vote 2 down vote accepted

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

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