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…
Best Answer
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.