Argument range for branch cut of $[-1,1]$?

Question

I read somewhere on this site that for a branch cut between $(-\infty, 0]$, the range of values for $\arg(z)$ is $[-\pi,\pi]$, while for a branch cut between $[0,\infty)$, the range of values is $[0,2\pi)$. Why is this so? Is this just a definition, or is there some mathematical reasoning to it? Does this apply to branch cuts along the imaginary axis? Also, for a branch cut that crosses the origin, such as $[-1,1]$ for $\sqrt{z^2-1}$, what should the approach be?

Edit: the question on stackexchange I was referring to is this one.

Answer 1

The ${\rm arg}$ function is multiple valued on $\dot{\mathbb C}$. In order to obtain a well defined and continuous (even smooth) real valued function on a slightly restricted domain one then makes the branch cuts. If the cut is made along the negative real axis one then defines $${\rm Arg}(z):={\rm the}\bigl({\rm arg}(z)\>\cap\>]{-\pi},\pi[\>\bigr)$$ as principal value of the argument, and if the cut is made along the positive real axis one can take $$\phi(z):={\rm the}\bigl({\rm arg}(z)\>\cap\>]0,2\pi[\>\bigr)$$ as smooth real representant of the argument. In both cases the chosen function jumps by $2\pi$ when you cross the cut. It would be pointless to make a cut along the negative real axis and then choosing $\>]0,2\pi[\>$ as range of the argument.

The cut along $[-1,1]$ is of no help for the argument function, but in other cases it is. The function $$f(z):=\sqrt{z^2-1}\qquad(?)$$ is double valued for all $z\ne\pm1$. But if you make a cut along $[-1,1]$ you can single out the branch that is $\>= +z+o(1)$ for $|z|\to\infty $, and will never get into trouble, i.e., obtain a single valued function $f$ defined on all of ${\mathbb C}\setminus[-1,1]$.

Answer 2

You have to first understand what the purpose of branch cuts are. It originates at the origin. The value of $\, \arg(0) \,$ is undefined, but it is worse than that. If you start at a point $\,z\,$ and loop around the origin one or more times and end up at the same point $\,z\,$, then the value of $\,\arg(z)\,$ varies monotonically continuously and ends up differing by a multiple of $\,2\pi\,$ from its original value, assuming no branch cut is encountered.

This kind of multi-valued behavior happens for $\, \sqrt{z} \,$ as well. One practical solution is to prevent looping around the origin using branch cuts. The two branch cuts you mention both perform this function well. But, a branch cut that crosses the origin and does not go to infinity does not perform this function, and hence useless for its intended purpose.

Answer 3

Branch cuts occasionally refer to the following two different concepts, which your question seems to mix together:

• A maximal subset of $\Bbb C$ on which a function can be defined while still being holomorphic (or at least continuous), e.g. $\Bbb C\setminus(-\infty,0]$ for $z^{1/2}$.
• A subset of $\Bbb C$ on which a function $\arg(z)$ (which satisfies certain properties, see below) can be defined.

The former ones can lie basically everywhere in the complex plane (see the examples here). However, they do not necessarily provide appropriate domains for defining $\arg(z)$, as is evident from your $[-1,1]$-example.

The function $\arg:C\to\Bbb R$ can only be defined on simply connected sets $C\subset \Bbb C$ which do not contain the origin. Usually, it is also required that $C$ is open and connected, which then is called a domain. Given such a domain $C$, the function $\arg(z)$ is uniquely determined up to an additive constant $2\pi k,k\in\Bbb Z$, e.g. by the following requirements:

• $\arg(z)$ is continuous on $C$,
• $|z|\exp(i \arg(z))=z$ for all $z\in C$.

So given an appropriate branch cut, you do not have much choice on how to choose $\arg(z)$. The proof of this usually involves topology.

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Examples

• Choosing $C=\Bbb C\setminus(-\infty,0]$ gives the range of $\arg(z)$ to be $[-\pi+2\pi k,\pi+2\pi k]$ for a $k\in\Bbb Z$ of your choice. Choosing $k=0$ gives the usual range $[-\pi,\pi]$.

• The following spiral branch cut can give $\arg(z)$ a range of $(-\infty,2\pi k)$ for a $k\in \Bbb Z$ of your choice.

• $C=\Bbb Z\setminus\{0\}$ is not simply connected, hence there does not exist a function $\arg(z)$ on it with the requried properties.