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In this function:

$$f(z) = \log(z-1) - \log(z+1)$$

it seems to me like there is only one option for the branch cut for this problem, between $z=1$ and $z=-1$ but I’m seeing other resources that indicate that there are several options.

On page 13 of this resource it indicates that there is more than one option for the branch cut.

Here’s my logic: - since $z=1$ and $z=-1$ are the only two branch points - and since we must draw branch cuts such that it is impossible to draw a circle around any one branch point without crossing the branch cut - and since $z=\infty$ is not a branch point - then it seems to me like the line connecting $z=1$ and $z=-1$ is the only option for the branch cut. But this resource indicates that there are these two options for branch cuts:
branch point options for <span class=$f(z)=\log(z+1)-\log(z-1)$"> But I don’t understand why the rightmost option is even an option. There’s no branch point at $z=\infty$

ViktorStein
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makansij
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    There are an infinite number of ways to cut the plane. A branch cut need not be a straight line. So any curves that originate at $\pm 1$ and meet at a commin point, including the point at infinity qualify. – Mark Viola Nov 26 '19 at 21:18
  • Thanks Mark Viola. And actually, I understand that. But Figure 2.5 (shown in my question) seems to indicate that the branch cut can be drawn extending from $z=1$ and $z=-1$, and not meeting at a common point. (unless they meet at infinity and it's just not shown in the figure). But, to be clear, they are not required to meet at $z=\infty$, because it's not a branch point? – makansij Nov 26 '19 at 22:25
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    The text above the figure does mention "We may also take the cuts as in Figure 2.5 (we may think of them as a straight line joining z = -1 to z = 1, which passes through $\infty$)". I've gotta say this sentence is confusing to me, since a straight line connecting $-1$ and $1$ can only be drawn between them, so I'm have a hard time seeing how the figure illustrates what they're saying. – makansij Nov 26 '19 at 22:28
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    They meet at "The Point at Infinity," which corresponds to the point $(0,0,1)$ on The Riemann Sphere. Are you familiar with The Riemann Sphere and The Point at Infinity? In the Extended Complex Numbers $\mathbb{C},\cup,{\infty}$ there is one point at infinity. So, any curve that beings at $z_1\in \mathbb{C}$ and ends at $\infty$ meets any curve that begins at $z_2 \in \mathbb{C}$ and ends at $\infty$. – Mark Viola Nov 26 '19 at 22:28

1 Answers1

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In the extended complex numbers, $\mathbb{C}\,\cup\,\{\infty\}$, the point at infinity is the stereographic projection from the point $(0,0,1)$ on the Riemann Sphere onto the complex plane $\mathbb{C}$.

Hence, any curve $C_1$ that begins at $z_1\in\mathbb{C}$ and ends at the point at $\infty$ (i.e., at $(0,0,1)$ on the sphere) joins at the point at $\infty$ ($(0,0,1)$ on the sphere) with any curve $C_2$ that begins at $z_2\in \mathbb{C}$ and ends at infinity.

As just one example, for $z_1=-1$ and $z_2=1$, choose the curve $C_1$ to be the set of points $\{z|\text{Im}(z)=0, \text{Re}(z)\le -1\}$ and choose the curve $C_2$ to be the set of points $\{z|\text{Im}(z)=0, \text{Re}(z)\ge 1\}$. These curves meet at the point at infinity (i.e., $(0,0,1)$ on the sphere).

That said, we could choose to cut the plane with the any curve that begins at $z=-1$ and ends at $z=1$. This curve need not be the straight line adjoining the branch points. Or we could have any curves, not only linear paths on the real axis), that begin at $\pm 1$ and terminate at the point at $\infty$.

Mark Viola
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  • okay, thanks a lot! this is starting to make sense. And when we say these are "valid branch cuts", what we mean is that if we draw any circle which does not intersect the branch cut, then we will not result in a multivalued function. However, I have seen some resources (on page 53: http://www1.spms.ntu.edu.sg/~ydchong/teaching/07_branch_cuts.pdf) test for multivaluedness, even after a valid branch cut has been drawn. So, if we draw the branch cuts as such, then do we still need to test for multivaluedness by drawing a circle? Or is guaranteed to be single valued? – makansij Nov 27 '19 at 03:13
  • Okay, now I think I really get it: Yes it is guaranteed to be single valued. However. each different choice of a principal Argument will result in a different branch cut. So, your choice of a principal Argument is essentially what determines which branch cut you use. Is my understanding accurate? – makansij Nov 27 '19 at 03:33
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    This Answer, this one, and this one might be helpful in understanding what is going on here. – Mark Viola Nov 27 '19 at 03:36
  • Thanks. But is my understanding accurate (see above)? – makansij Nov 27 '19 at 20:36
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    @nundo You're welcome. I suggest that one analyze one branch point at a time. Choose a branch cut for that branch point so that in the "cut" plane, the function of interest is single-valued. Then, analyze the second branch point, etc. – Mark Viola Nov 27 '19 at 23:42
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    @nundo Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Dec 01 '19 at 00:57
  • It's already perfect! sorry for not accepting it earlier! – makansij Dec 01 '19 at 20:03
  • @MarkViola When we make a branch cut do we exclude the blue points on the figure or do we include them? – moli May 26 '21 at 14:07