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?
Is the local inverse of an analytical function locally analytical as well?
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complex-analysis
exponentiation
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0@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
1 Answers
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See the Lagrange inversion theorem: "$g$ is analytic at the point $b=f(a)$".