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I am trying to find a definitions which link the concept of branches, branch cuts and branch points. Relating to this I have a question concerning the nature of branch cuts: Do they necessary have to be lines? i.e. could they be strips - or any other shaped region (along as it keeps the branch single-valued and continuous)?

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    $\begingroup$ Anything you like. $\endgroup$ – Marja Aug 5 '17 at 12:30
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    $\begingroup$ @Quantum: It's customary to take branch cuts to be "cuts", piecewise-smooth curves joining pairs of branch points, but as Marja says cuts needn't be lines. For example, there exist branches of $\log$ on the complement of (the closure of) a logarithmic spiral. (The imaginary part of such a branch is unbounded.) $\endgroup$ – Andrew D. Hwang Aug 5 '17 at 12:57
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    $\begingroup$ @AndrewD.Hwang My concern is that under this interpretation only very special branches have branch cuts, i.e. those whose complement of their domain is 'piecewise-smooth curve'. For a branch defined on a much smaller domain - what then is the (large) complement of this domain called? $\endgroup$ – Quantum spaghettification Aug 5 '17 at 13:26
  • $\begingroup$ In my experience, one usually speaks of "branch cuts" only when one has a multi-valued function defined over the complex plane (i.e., a Riemann surface and a branched covering to $\mathbf{C}$), and wants to divide the Riemann surface into graphs ("branches"). Do you have an example in mind whose domain has large complement? $\endgroup$ – Andrew D. Hwang Aug 5 '17 at 13:49
  • $\begingroup$ @AndrewD.Hwang I don't but, but I could easily make one, e.g. define a branch of $z^{1/2}$ to be $f_1=\sqrt{z}$ defined for only $0\le \theta \lt \pi/4$. This (at least I think) would technically count as a branch and has a large complement where the function is not defined. $\endgroup$ – Quantum spaghettification Aug 5 '17 at 14:04
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Note I am the OP. I have been looking the wrong way wrong. The following is a definition of a branch cut (from these notes):

A branch cut for a multifunction $f$ is a curve in the plane on whose complement we can pick a holomorphic branch of $f$. Thus a branch cut must contain all the branch points.

As you can see from this definition the branch cut defines the holomorphic branch not the other way round. Thus it is true that not all branches of a function have 'branch cuts' but it is true that all branch cuts are associated with a holomorphic branch.

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    $\begingroup$ Your answer doesn't explain what needs to be explained. Define the $\epsilon$-almost neighborhoods of $a$ as the simply connected open sets contained in $0 < |z-a|< \epsilon$. If $f$ is analytic on some almost neighborhood of $a$ and for every $\epsilon$-almost neighborhood $U$ of $a$, $f$ has an analytic continuation which is analytic on $U$, but none of them is analytic on $0 < |z-a|< \epsilon$, then $a$ is a branch point of $f$. $\endgroup$ – reuns Aug 6 '17 at 3:44
  • $\begingroup$ In that case (the analytic continuation of) $g(z) = f(a+e^{-z})$ is analytic for $\Re(z)$ large enough. And $a$ is a branch point of $f$ iff $g$ is not $2i\pi$ periodic. Then, you can define $g(z+2i\pi)-g(z)$ which is the functional equation of the branch point. $\endgroup$ – reuns Aug 6 '17 at 3:50

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