Isn't there a general way to find the branch points of a complex function? I mean, one cannot just go around every point to see if the function is multi-valued. Isn't there some sort of classification letting know how to deal with functions of a special type?

Perhaps I'm asking this because I haven't digest the concept of branch point well. I've been reading that a branch point is a singularity, but don't understand the difference between a branch point and a singular point which is not a branch point! I know that the function is not continuous at a branch point; so it can't be analytic also, right?

  • $\begingroup$ of what function ? it depends on if you have an expression for it on some disk, or for the whole complex plane, or if you only have a complicated power series on some disk, or if you only defined it by some functional/differential equation, etc. $\endgroup$ – reuns Apr 6 '16 at 18:02
  • $\begingroup$ and in "singular point" the key-word is "point". $f(z)$ has a singular point at $a$ if it is holomorphic on some neighborhood of $a$ except at $a$. look at $f(z) = \ln(z)$ or $f(z) = z^{1/2}$, it is not holomorphic on any pointed neighborhood of $0$, hence $0$ is not a singular point. and it has a branch cut because it is discontinuous on some line (but it is holomorphic on the neighborhood of that line). $\endgroup$ – reuns Apr 6 '16 at 18:06
  • $\begingroup$ @user1952009 So you mean the function is holomorphic on that point, but not on its neighborhood? And how do I find out if the function is not holomorphic on some neighborhood of a point? $\endgroup$ – Simorq Apr 6 '16 at 18:31
  • $\begingroup$ the function is not holomorphic at that point but is holomorphic on its neighborhood. for example $f(z) = 1/z$ or $g(z) = e^{1/z}$ both are holomorphic on $0< |z| < 1$ but not at $z=0$ : hence $0$ is a singular point of those functions. now if you take $h(z) = z^{1/2}$ it is NOT holomorphic on $0 < |z| < \epsilon$ for any $\epsilon > 0$ hence $0$ is NOT a singular point (it is a branching point of some branch cut) $\endgroup$ – reuns Apr 6 '16 at 19:02
  • $\begingroup$ and $h(z) = z^{1/2}$ is not holomorphic on $0 < |z| < \epsilon$ because if you choose the usual branch of the logarithm, hence $z = r e^{ i \theta}$ with $\theta \in ]-\pi,\pi]$ you get $h(r e^{i \theta}) = \sqrt{r} e^{i \theta / 2}$ which is discontinuous around $\theta = \pi$ $\endgroup$ – reuns Apr 6 '16 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.