Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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) – qwertyuio Apr 12 '13 at 11:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.