[Complex analysis] Contradiction in the definition of a branch

[fatpotato]

Homework Statement

Using the definition of a branch to show that $Arg z$ is a branch of $\arg z$

Relevant Equations

Definition of a branch, definition of principal argument $Arg z$, definition of argument function $\arg z$

I find the following definition in my complex analysis book :

Definition : $F(z)$ is said to be a branch of a multiple-valued function $f(z)$ in a domain $D$ if $F(z)$ is single-valued and continuous in $D$ and has the property that, for each $z$ in $D$, the value $F(z)$ is one of the values of $f(z)$.

I am now assuming that the choosen domain is in fact $\mathbb{C}$, an open and connected set. Now, the first function given as an example for this definition is $Arg z$, the principal argument of $z$ (is there a latex command for this function?) that relates to $\arg z$ with $Arg z = \arg_{-\pi} z \in (-\pi;\pi]$ where $Arg z$ is simply the argument of the complex number $z \neq 0$ that can take any value in the interval $(-\pi;\pi]$.

Acording to the book, $Arg z$ is a branch of $\arg z$ because (considering $z \neq 0$):

1. It is single-valued on $\mathbb{C}$ - I'm ok with this
2. $Arg z$ is one of the values of $\arg z$ - I'm still ok with this
3. It is continuous on $\mathbb{C}$ - I'm not okay with this!

How can an inherently discontinuous function be a branch? Everytime the number $z$ will cross the branch cut on the negative real line, there will be a jump from $-\pi$ to $\pi$, will it not? The book even makes a point of insisting that there will always be a jump of $2\pi$ when considering the argument of a complex number. This confuses me greatly, and I need branches in the next chapter since they are used to define the behaviour of the complex logarithm.

Anyone has an idea?

Edit: missing sentence, add condition $z\neq 0$

 

[FactChecker]

Someone, either the author of the book or you, are not taking the point z=0 seriously enough. The function is not even defined there, so $\mathbb{C}$ is not the domain. To define the principle branch, the points, $r$, along the negative real axis is not considered to be in the neighborhood of the complex values $r-\epsilon i, \epsilon \gt 0$. So there is no continuity across the negative real axis. In fact, any branch for the argument function must have some slit in $\mathbb{C}$ from $z=0$ to $\infty$ across which there is no continuity.

 

[fatpotato]

I was unfortunately not rigorous enough. The book mentionned how $\arg z$ is not defined at $z=0$, which is what I tried to convey by specifying that $z \neq 0$. I should have mentionned that, for the given example, the domain $D$ is $\mathbb{C} \setminus \{0\}$.

Regarding your message, if there is no continuity across the negative real axis, this should mean that $Arg z$ is not a branch of $\arg z$. Is the book's definition erroneous?

 

[FactChecker]

You should not consider the domain for arg(z) to be $\mathbb{C}-\{0\}$. If you follow a path that winds around $z=0$, you need to think of there being several surfaces winding around where the arguments are continuous and unbounded. That will be addressed in greater detail later in a complex analysis class.

 

[vela]

I am now assuming that the chosen domain is in fact $\mathbb{C}$, an open and connected set.

I think your assumption about the domain is incorrect. It should be $\mathbb{C}$ minus the branch point and branch cut, in this case, the non-positive real axis.

 

[fatpotato]

You should not consider the domain for arg(z) to be C−{0}. If you follow a path that winds around z=0, you need to think of there being several surfaces winding around where the arguments are continuous and unbounded. That will be addressed in greater detail later in a complex analysis class.

What should the domain of arg⁡z be? If I go back to the definition of this function I have :

$$ \arg z =
\left\{
\begin{array}{ll}
\arctan \frac{y}{x} + \frac{\pi}{2}(1 - sgn(x)) & \mbox{if } x \neq 0 \\
\frac{\pi}{2}sgn(y) & \mbox{if } x=0, y\neq 0 \\
undefined & \mbox{if} x=y=0
\end{array}
\right\} $$

Where $sgn(x)$ function evaluates as $1$ if $x$ is positive, $-1$ if $x$ is negative and $0$ if $x$ is zero.
Does this mean that if I want Argz to be a branch of arg⁡z, I should further restrict the domain?

I think your assumption about the domain is incorrect. It should be C minus the branch point and branch cut, in this case, the non-positive real axis.

But considering C minus the branch cut, Argz would never take the value π? How can we exclude the negative real axis?

Also, I am sorry if this has been answered and flew over my head, but I still see Argz as a discontinuous function that could not possibly fill the criteria for a branch, therefore contradicting the book's example of Argz being a branch of arg⁡z. Could someone address this specific point?

Edit: add definition for $sgn(x)$

 

[aheight]

With regards to multi-valued functions, everything begins with $\arg(z)$ so good idea to have a good grasp of it. First, it's an infinite helical coil winding over the complex plane. But to get some practical handle on it, we can just look at a few windings keeping in mind they wind infinitely upward and downward. The plot shows three winding for $z=re^{it}$ with $-3\pi\lt t\leq -\pi$ red, $-\pi\lt t\leq \pi$ blue, and $\pi\lt t\leq 3 \pi$ green. And in the figure, there are three points on this 3-surface plot over every value of z. But in reality, since the coil are infinite, there are infinitely many but just for now, consider just the three in the plot. Now, this is the crucial question about branching: How can we "excise" a continuous "piece" of this infinite coil that is single-valued for all values of z except 0 and continuous in the interval $\theta\lt t\leq 2\pi+\theta$?" That's easy: take any $\theta\lt t\leq 2\pi+\theta$ section of it. Right? By convention then we define the "principal" sheet as the section $-\pi\lt t\leq \pi$ and we name this section $\text{Arg}(z)$. And keep in mind branches of multivalued functions are by their nature "discontinuous" on their branch-cuts. With regards to $\text{Arg}(z)$ the discontinuity is along the negative real axis.

 

[fatpotato]

Thank you very much for the time you took writing your answer and making this graph.

I agree that the discontinuity is along the negative real axis regarding $\text{Arg}\,z$, so the textbook definition of a branch, that I wrote in my initial post, is incorrect since it requires a branch to be continuous.

 

[WWGD]

Thank you very much for the time you took writing your answer and making this graph.

I agree that the discontinuity is along the negative real axis regarding $\text{Arg}\,z$, so the textbook definition of a branch, that I wrote in my initial post, is incorrect since it requires a branch to be continuous.

Sorry for the necropost. Not sure what you mean by " continuous " here, but the Region $\mathbb C - (-\infty,0)$ is certainly connected and even simply-connected.