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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1yes, 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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0See http://en.wikipedia.org/wiki/Lagrange_inversion_theorem. – 2011-03-10
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0@yoyo: you mean yes for the example but in general I have to... – 2011-03-10
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0...use the formula linked by @joriki and prove the convergence on a case-to-case basis? – 2011-03-10
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1The 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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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 $b=f(a)$".