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Both Wikipedia and MathWorld (here and here) seem to place some imporantance on saying, but not elaborating on

It should be noted that the endpoints of branch cuts are not necessarily branch points.

When would it make sense to have a branch cut that ends at a non-branch point?

Is it just because one can concoct spurious cases such as $$ f(z) = \begin{cases} 1 & |z| < 1 \\ 2i & |\Im(z)| > 2 \\ z & \text{otherwise} \end{cases}$$ with a "branch" cut with no ends along the unit circle, and another one going from infinity to infinity? Or is there more to it?

hmakholm left over Monica
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  • @Random: Which cut are you using there? For example, with the one Wolfram uses, the branch cut seems to pass through the point at infinity rather than end there. – hmakholm left over Monica Sep 13 '15 at 02:59
  • I think you're correct. I was incorrectly interpreting it as two separate branch cuts ending at infinity and not as a single branch cut passing through the point at infinity and connecting $i$ with $-i$. – Random Variable Sep 13 '15 at 03:51

1 Answers1

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Hmm, here is one arguable case:

The lacunary function $f(z)=\sum_{n=0}^{\infty} z^{n!} $ is defined on the unit disc and has singularities at every point on the unit circle. Now, the function $$ g(z) = f(e^{-\sqrt z}) $$ where the square root is the usual principal square root, is defined on the complex plane except for the (closed) negative real axis.

Then one might consider the negative real axis to be a branch cut for $g$. Or perhaps not; that depends on exactly what one takes "branch cut" to mean and how seriously one takes the requirement (in both Wikipedia's and Wolfram's definitions) that the function must be "multi-valued".

In any case its endpoint at $0$ is definitely not a branch point: one cannot continue the function analytically along any closed curve around $0$.

Alternatively, and less dramatically, one may consider $$ h(z) = \sum_{n=1}^\infty \frac{\sqrt{z+1/n}}{2^{-n}} $$ where again all of the square roots are principal ones. Here the discontinuity along the negative real axis is actually a bona fide branch cut at all except countably many points -- but still the endpoint at $0$ is not a branch point.

hmakholm left over Monica
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  • Concerning your last sentence, wouldn't that make $0$ not a branch point of the logarithm by the same reasoning? But in the wiki article it is claimed that $0$ is a (logarithmic, of course!) branch point of the logarithm. I think you can only argue that your $g$ does not have an algebraic branch point at $0$. – guest Sep 12 '15 at 21:51
  • @guest: I was expressing myself sloppily, and probably shouldn't have mentioned curves. My point is that for a branch point you can continue the function analytically around it and fail to match up. More rigorously expressed I'd say that for $z_0$ to be a branch point there should be analytic functions $f_1$ and $f_2$ on open connected domains whose union is a punctured disc around $z_0$, and such that $f_1$ and $f_2$ agree on a proper subset of the intersection of their domains (and $f_1$ agrees with the original $f$ on some open set). – hmakholm left over Monica Sep 12 '15 at 22:03