# Do branch cuts need to be lines?

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)?

• Anything you like. – Marja Aug 5 '17 at 12:30
• @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.) – Andrew D. Hwang Aug 5 '17 at 12:57
• @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? – Quantum spaghettification Aug 5 '17 at 13:26
• 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? – Andrew D. Hwang Aug 5 '17 at 13:49
• @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. – Quantum spaghettification Aug 5 '17 at 14:04

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.
• 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 $f$. – reuns Aug 6 '17 at 3:44
• 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. – reuns Aug 6 '17 at 3:50