If $f:D\to D'$, with $D, D'$ open subsets of $\mathbb{C}$, is a complex analytic invertible function with non-zero derviative, it's easy to see that $f^{-1}:D'\to D$ is analytic too. Indeed complex analytic functions are just holomorphic functions and $\exists\ \frac{df^{-1}}{dz}=(\frac{df}{dz})^{-1}$.

Now if $f:I\to I'$, with $I, I'$ intervals of $\mathbb{R}$, is a real analytic invertible function with non-zero derivative, is it true that $f^{-1}:I'\to I$ is analytic?

One may extend $f$ to a complex analytic function, but I don't know if this one is still invertible...


1 Answer 1


Near any point $a\in I$, $f$ is given by a power series in $x-a$, which can be interpreted as a complex power series, which also converges in an open neighbourhood of $a$, hence describes a holomorphic function with nonzero derivative at $a$, hence is locally invertible, hence the local inverse holomorphic, is given by a complex power series and this power series must have real coefficients by symmetry.

  • $\begingroup$ Thank you very much. So is it true that an holomorphic function with non-zero derivative is locally invertible? (exactly as a rreal differentiable function with non-zero derivative) $\endgroup$
    – qwertyuio
    Apr 12, 2013 at 11:29
  • $\begingroup$ I don't really like the proof using the theorem that all complex functions that are differentiable in an open disk are analytic in that disc. It just blindly applies calculations without showing me what's going on which is quite advanced. You can however find the formula for computing the inverse of a power series under certain assumptions. Then you can check for yourself that when you take the original power series and substitute the calculated inverse power series for x, you get the identity function in a disk of nonzero size. $\endgroup$
    – Timothy
    Jun 9, 2020 at 18:57

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