0
$\begingroup$

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 ?

2 Answers 2

2

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.

  • 0
    That last condition is wrong. As a clean example, consider the function $f(z) = \sqrt{(z+1)^2+1} + \sqrt{(z-1)^2+1},$ which has four branch points at $z=\pm1\pm i$ (with independent signs), and for which it is possible to join up the branch points [pairwise](http://i.stack.imgur.com/Fwdg8.png) in a way that a curve can enclose two out of the four branch points and not intersect any branch cut.2017-11-25
5

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,

  • 0
    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.2012-12-27