Consider the function $$f(w)=\frac{e^{2\pi iw}-1}{w+1}$$
I guess that this function is surjective except at integer $w$. My argument is:
$\frac{1}{w+1}$ is surjective on the whole complex plane.
$e^{2\pi iw}-1$ is $1$ periodic.
Therefore the product of these two functions is surjective on the whole complex plane except the zeroes of $e^{2\pi iw}-1$, i.e. integer $w$.
Is this argument strong enough? Is it supported by any theorem?
Moreover, I would like to obtain the inverse function of $1/f(w)$.
I guess the inverse function is single-valued except at infinity, and therefore has an infinite radius of convergence.
Following the notations on Wikipedia page of Lagrange Inversion Theorem, I obtained $$g_n=\sum^n_{r=0}(-1)^{n-r}(2\pi ni)^{n-1}$$
Is this correct? I think such simple function should have a simpler inverse.
Thanks in advance.
p.s. If it turns out that the inverse of $f(w)$ is even more beautiful, please kindly provide it in your answer.
No, the inverse is not single-valued. The solutions of $1/f(w) = z$ are $$ w = {\frac {i/2{\rm W} \left(-2\,i\pi\,z{{\rm e}^{-2\,i\pi\,z}}\right)}{ \pi}}-z-1 $$ where $W$ is the Lambert W function. This is multivalued, with infinitely many branches.