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I am learning some complex analysis and I still have problem with branch cuts.

I tried answering (to myself) what is the range of the argument,$\theta$, in $$z^{\frac{1}{n}}=\sqrt{n}e^{i\frac{\theta}{n}}$$ in different branches and how does the branch cuts looks like.

Can someone please help me with understanding how the range of the arguments is chosen, and given a range, how to draw the corresponding branch cut ?

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2 Answers 2

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This is explained in Wikipedia :

  • case $n=2$ : draw the two branches $\ z\mapsto z^{1/2}$ and $\ z\mapsto z^{1/2}e^{2\pi i/2}$ (we display the imaginary part only for $\ x+iy\,\mapsto z:=\Im (x+iy)^{1/2}\,$) n=2

  • case $n=3$ : draw the three branches $\ z\mapsto z^{1/3}$ , $\ z\mapsto z^{1/3}e^{2\pi i/3}$ and $\ z\mapsto z^{1/3}e^{4\pi i/3}$

n=3

  • case $n=4$ : draw the four branches $\ z\mapsto z^{1/4}$ , $\ z\mapsto z^{1/4}e^{2\pi i/4}$ , $\ z\mapsto z^{1/4}e^{4\pi i/4}$ and $\ z\mapsto z^{1/4}e^{6\pi i/4}$ n=4

Hoping this helped to imagine to generalize,

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  • $\begingroup$ Thanks for the answer. How, for example, is $z^{\frac{1}{4}}$ defined ? where do you take $\theta$ to be in ? $\endgroup$
    – Belgi
    Dec 27, 2012 at 0:15
  • $\begingroup$ Are those drawings of the branch cuts ? What I studied is that the branch cut is something like whats described in Ted's answer $\endgroup$
    – Belgi
    Dec 27, 2012 at 0:17
  • $\begingroup$ For $z:=\rho e^{i\theta}$ you may for example take $\theta\in (-\pi,\pi)$ (in fact it is the choice of the software!). Here we are drawing the branches (the imaginary parts are visualized), the cut is at $\theta=\pi$ (corresponding to $-\pi$). $\endgroup$
    – Raymond Manzoni
    Dec 27, 2012 at 0:23
  • $\begingroup$ And how did you know that you get the (all) the other branches by multiplying by $e^{2\pi i/4}$ ? $\endgroup$
    – Belgi
    Dec 27, 2012 at 0:26
  • $\begingroup$ In general you get the next branche by multiplying by $e^{2\pi i/n}$. Note that you will have a repetition when considering the $n$-th multiplication since you'll get a total multiplicative factor of $e^{n 2\pi i/n}=1$. There are exactly $n$ different branches if $n$ is integer. $\endgroup$
    – Raymond Manzoni
    Dec 27, 2012 at 0:31
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How you choose the range of the argument, and the corresponding branch cuts, depends on what you are doing. In a particular problem, you may have some existing "standard" values that you need to agree with (for example, if $z$ is a positive real number, then $\sqrt{z}$ may need to be the positive square root, or $\log z$ may need to be the real logarithm). These conditions will constrain your choices.

Even after you decide the range of the argument, branch cuts are pretty flexible. Generally, they should be curves connecting all the branch points, and they shouldn't go through any points you're interested in for your problem, but other than that, you can choose whatever. For $z^{1/n}$, the only branch points are 0 and $\infty$ (since all other numbers have $n$ distinct $n$th roots), so a branch cut can be any curve joining 0 to $\infty$.

Edit from the comments below because I think it's important enough to be in the answer:

The precise condition for choosing branch cuts (I believe -- I'm not an expert at this stuff), is that they must be chosen so that any closed curve in $\mathbb{C}$ not intersecting the branch cuts will enclose either all or none of the branch points.

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  • $\begingroup$ Thans for the answser! so to deduce what a branch cut can be I should find the branch points and the branch cut is any curve joining all of them and goes to $\infty$ ? $\endgroup$
    – Belgi
    Dec 27, 2012 at 0:14
  • $\begingroup$ The branch cut should only go to $\infty$ if $\infty$ is a branch point. There's nothing special about $\infty$; you treat it like any other branch point. For example, if the function were $\sqrt{1 - (1/z)}$, then $\infty$ is not a branch point; the branch points would only be 0 and 1. You could choose a straight line joining 0 and 1 as your branch cut. $\endgroup$
    – Ted
    Dec 27, 2012 at 0:18
  • $\begingroup$ How can I tell if $\infty$ is a branch point ? it isn't actually a point in $\mathbb{C}...$ . In general I should find the branch points and the branch cut is any curve joining all of them ? (I removed the requirement to go to $\infty$) thanks again for the help! $\endgroup$
    – Belgi
    Dec 27, 2012 at 0:21
  • $\begingroup$ Intuitively, a branch point is wherever the function behaves "differently" from elsewhere, either because there are fewer "choices" than usual or because it's undefined. For example, most complex numbers have 2 square roots, but 0 and $\infty$ only have 1, so they are the branch points for $\sqrt{z}$. I know this isn't a precise statement; a precise statement can be found in the theory of compact Riemann surfaces (and the reason we include $\infty$ is so that our space becomes compact). $\endgroup$
    – Ted
    Dec 27, 2012 at 0:32
  • $\begingroup$ Thanks again! I actually prefer at this moment to get the intuition and not the precise definition so this helps a lot. In regards the the question in my previews comment: Is that the way I can construct a branch cut ? (any curve that joins all the branch cuts ?) $\endgroup$
    – Belgi
    Dec 27, 2012 at 0:34

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