1

If $w = z^n$, then given any point $w = r e^{i \theta}$ with $r > 0, \theta \in \mathbb{R}$, there are $n$ distinct points of the $z$-plane with $w$ as their image:

$$z = \sqrt[n]{r} e^{i(\theta + 2 \pi k)/n} \ \text{ for } \ k = 0, 1,..., n-1$$

To define an inverse of $w$, we need the map to be univalent. The textbook I'm following (Silverman) states that to get the maximal domain in which $w$ is univalent, we must have $c < \arg(z) < c + 2 \pi/n \ $ for some real number $c$.

If we arbitrarily choose $c = 0$, we get $0 < \arg(z) < 2 \pi/n \implies 0 < \arg(w) < 2 \pi$, so that we have a branch cut along the positive real axis in the $w$-plane.

My question is, why can we not choose $0 \le \arg(z) < 2 \pi/n$?

They seem to be implying that it will break the univalence, but I'm having trouble seeing where this occurs. As far as I can tell, neither the injectivity nor analyticity would be broke if we included $\arg(z) = 0$ as an option, and in this case, the image of $w$ would be the entire $w$-plane, so that no branch cuts occur.

My thought process on the injectivity was that, if $w = z^n= r$, a positive real number, then

$$z = \sqrt[n]{r} e^{2 \pi k i/n} \ \text{ for k= 0, 1,..., n-1 }$$ Are the $n$ distinct points mapping $w$ maps to $r$. If we restrict ourselves to $0 \le \arg(z) < 2 \pi/n$, then $z = \sqrt[n]{r}$ is the only option, implying a one-to-one mapping.

infinitylord
  • 5,146
  • Look at J.Loreaux answer there : https://math.stackexchange.com/questions/188359/when-does-a-complex-function-have-a-square-root/188433 – zwim Dec 17 '17 at 04:18
  • Perhaps Silverman wants a function whose domain is open – ziggurism Dec 17 '17 at 04:25
  • @zwim: Thanks for the reference. I am still having confusion over the problem stated, but J. Loreaux's statement that "an nth root function will exists iff the complex logarithm exists" was enlightening, for (perhaps ironically) I understand why the inequality cannot be strict for logarithms, which would cause the inequality to not be strict for the nth root. – infinitylord Dec 17 '17 at 04:26
  • @ziggurism: Actually, that is a very good point. Silverman works almost exclusively with open domains (indeed, the very definition of domain in this book is an open connected set). Is that to say if we considered closed connected sets as domains, we could get away with the nth root being defined without branch cuts? – infinitylord Dec 17 '17 at 04:27
  • Analytic makes sense only on open sets, that's all. The image of your open set is $\mathbb{C} \setminus [0,a \infty)$, its closure is $\mathbb{C}$. – reuns Dec 17 '17 at 04:34
  • @infinitylord just to point out that your proposed domain is neither open nor closed. Considering closed domains doesn't get you anything, and isn't the maximal domain. – ziggurism Dec 17 '17 at 04:53
  • @infinitylord for the record, I think you are correct that you can increase the domain in the way described in your question. – ziggurism Dec 17 '17 at 04:55
  • @ziggurism : How do you increase the domain without making the function (generally) discontinuous, hence non-differentiable? – Eric Towers Dec 17 '17 at 05:00
  • @EricTowers I cannot vouch for the differentiability of the function on this larger domain. But I think continuity should be okay? – ziggurism Dec 17 '17 at 05:11
  • 1
    @ziggurism continuity in the argument, but not in the complex topology (indeed analytic on some open minus a line and continuous on the whole open implies analytic on the whole open) – reuns Dec 17 '17 at 05:15
  • @reuns I'm sorry of course you guys are right. $z\mapsto z^{1/n}$ is not continuous on all of $\mathbb{C}$ (which is what the domain would have to be to surject onto ${0\leq \theta< 2\pi/n}$). The best we can say is it is bijective with infinitylord's choice of domain. – ziggurism Dec 17 '17 at 05:23
  • Thank you all, this has been an instructive read. Indeed, the bijectivity is the component I was focused on when thinking of the domain. I saw no reason to believe analyticity (or continuity, for that matter) would be compromised, but that at least explains why this domain fails to produce an inverse. – infinitylord Dec 17 '17 at 05:28

1 Answers1

1

If you "fill in" the branch cuts with a choice of sheet, then the resulting function on $\mathbb{C}$ isn't even continuous (in general) across the cut, much less holomorphic. If the branch cut is deleted from the domain, holomorphicity is retained.

Eric Towers
  • 70,953
  • 1
    Could you expound on why the function wouldn't be continuous without the cut? In some sense I can see why, as approaching the positive real line from above and below would produce different limiting values in the inverse function. However, I am having trouble seeing why this situation differs when we introduce the branch cut. Is it nonsensical to talk about the limiting behavior at a cut (in the sense that it is nonsensical to talk about a functions limiting behavior at a point outside of its domain)? – infinitylord Dec 17 '17 at 05:33
  • As $w$ descends past the real axis from above, the argument of its image decreases to zero then jumps to $2\pi/n$. This is not continuous. More generally, if it were continuous, the preimage of an open set is an open set. The preimage of your filled-in function of a small open disk centered at $1$ is two half disks, one including its diameter and one not. The component including its diameter is not open, so the map is not continuous. – Eric Towers Dec 17 '17 at 16:57