# About the inverse of a simple function

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:

1. $\frac{1}{w+1}$ is surjective on the whole complex plane.

2. $e^{2\pi iw}-1$ is $1$ periodic.

3. 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.

p.s. If it turns out that the inverse of $f(w)$ is even more beautiful, please kindly provide it in your answer.
• What does "surjective except at integer w" mean? It is certainly not true that the product of surjective functions is surjective. $f(x)=x$ is surjective on $\mathbf R,$ but $f\cdot f$ is not. – saulspatz Mar 30 '18 at 4:30
• $1/(w + 1)$ is not surjective (unless you're dealing with the Riemann sphere rather than the complex plane). And there's no reason to think the product of a surjective function and a periodic function is surjective. – Robert Israel Mar 30 '18 at 4:32
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.
• I never understand downvotes to good answers. $+1$ – Claude Leibovici Mar 30 '18 at 5:45