Conformal points, branch points, and analyticity of the mapping $z=w+e^w$? Consider the mapping of the $z$-plane to the $w$-plane given by $$z=w+e^w$$
There is no closed-form $w(z)$ that matches the map, but we can still discuss conformity.
$$\frac{dz}{dw}=1+e^w \implies \frac{dw}{dz} = \frac{1}{1+e^w}$$
It appears that $w$ is differentiable and thus analytic where $e^w \ne -1$. So the $w$ mapping fails to conform where $w_n = (2n+1)\pi i$. On the $z$ plane, that corresponds to where $z_n=w_n+e^{w_n}$ or
$$z_n=-1 + (2n+1)\pi i$$
But I was instructed that this mapping has two branch points on the $z$-plane, $z= -1\pm \pi i$. And, barring placement of the branch cut, I was instructed that the mapping of the $w$-plane is conformal all $z$ except for those two branch points (and their cuts). 
Where does this come from? These two points are in the set I found, but if anything, it seems like I have infinitely many $z_n$ that could be contenders for the branch point. Without knowing $w(z)$, how can I actually determine where the branch points are?

Additional details: It may be important that I was mapping the two rays parametrized below, which were supposed to run along the branch cuts and end at the two branch points.
$$z=x\pm i\pi \quad\quad x\in(-\infty,-1]$$
 A: Let $f(w) = w + \exp(w)$
Depending on how you make your branch cuts (which is completely arbitrary, my favorite branch cuts are usually spirals), you can make all the branch points appear on your "map", or just as few as only one of them.
If your instructor has in mind horizontal branch cuts going to the left, then you get countably many branches each with two branch points.
Since you want to cut $z$ to the left of the branch points you have to cut $w$ along the preimage of those halflines. 
$f(x + (2k-1)i\pi) = x + (2k-1)i\pi - \exp(x)$. $x \mapsto x - \exp(x)$ increases from $- \infty$ to $-1$ (at x=$0$) then decreases to $-\infty$, hence you have to cut $w$ along the lines $\Im(w) = (2k-1)\pi$.
By restricting $f$ to a horizontal strip $U_k = \{w \mid (2k-1)\pi < \Im (w) < (2k+1)\pi \}$, you get a bijection from $U_k$ to $\Bbb C$ with the two half lines $\{z \mid \Re(z) \le -1 \land \Im(z) = (2k\pm 1)\pi \}$ removed.
Then the inverse of those restrictions are "branches" with two branch points. But there are infinitely many of them.
You could also decide to make the branch cuts go to the right (or make them vertical, or whatever), so you have to cut up the $w$s along the preimages of the other halflines. This time the picture is not as simple, you have to cut $w$ along infinitely many "c" shaped curves, with infinitely many "c" stacked on top of each other.
Then you get infinitely many branches with only one branch point (corresponding to the regions inside the "c" curves), plus one branch with infinitely many branch points (corresponding to the big region on the left of all the curves)
