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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.

$$z=x\pm i\pi \quad\quad x\in(-\infty,-1]$$

zahbaz
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  • "Where does this come from?" -- it comes from the mind of the person who told you that. Sorry, mind readers are in short supply here. A possible interpretation is that if one runs branch cuts vertically, then taking care of $-1 \pm \pi i$ automatically takes care of the rest, since they are covered by the cuts. –  Oct 11 '15 at 05:09
  • @MiceElf We seem to agree then. So all of those $z_n$ are branch points, right? I considered running the branch cuts vertically (up for $-1+\pi i$ and above and down for $-1-\pi i$ and below) like you describe, but this instructor described the problem with the branch cuts horizontal and claimed that there were only two branch points. I just don't see that. – zahbaz Oct 11 '15 at 05:51

1 Answers1

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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)

mercio
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