4

I know this is the most typical example of branches and I think I don't get the concept... Could you help me by giving a detailed development leading to all the required branches? It'd help me understanding more complicated examples...

Thank you very much, this concept really is hard to understand for me...

dustin
  • 8,559
Dollydot
  • 205
  • Does this help post help: http://math.stackexchange.com/questions/988828/how-to-find-the-branch-points-and-cut/992857#992857 – dustin Nov 04 '14 at 12:33
  • I had seen this post right before, it does make it a little less weird to me but doesn't really enlight the whole thing, I see what the points are and why they motivate this cut, but then I'm stuck... – Dollydot Nov 04 '14 at 12:49
  • Alright, that's much simpler, now for my example, the cut is there because if you start at a point z and turn around the origin then the argument varies by multiples of $2\pi$ right? – Dollydot Nov 04 '14 at 13:37
  • But then what exactly is a branch from there? – Dollydot Nov 04 '14 at 13:38
  • Are they just the one-valued way of defining log on the required set? Is it right to say the branches for me are just $f_n(z)=log(r)+i\theta+2ni\pi$ where $z=re^{i\theta}$ and $\theta\in(-\pi,\pi)$? How can I explain but by saying that the argument of $log(z)$ can vary by $2n\pi$ and so to avoid variations we restrict to intervals of size $2\pi$? – Dollydot Nov 04 '14 at 13:49

1 Answers1

3

The purpose of a branch cut or cuts is take a multivalued function $f(z)$ into a single analytic branch of $f(z)$. In the example of $f(z) = \log(z)$, lets consider moving around $z = 0$, from a point on the unit circle. Now, the branch point in this case is at $z=0$.

Let $z = e^{i\theta}$. Now if we start at $z = 1$ so $\theta = 0$, then $f(1) = \log(1) = 0$. Let's traverse around $z = 0$ by $2\pi$; that is, $z = e^{2\pi i}$. $$ f(e^{2\pi i}) = \log(e^{2\pi i}) = \log(1) + 2\pi i = 2\pi i $$ Therefore, as we encircle $z = 0$ by $f(1) = 0\mapsto f(e^{2\pi i}) = 2\pi$. Therefore, $f(z)$ is multivalued. In order to prevent this, we must stop the wrapping around $z = 0$ by a full revolution or $2\pi$. In my encounter, the principal value is general defined as $\theta\in(-\pi, \pi]$ but I have seen it defined as $\theta\in(0, 2\pi]$. If we use the first definition of principal value, we would define our branch cut as $(-\infty, 0]$ as you did in your question but we could also define it as $[0, \infty)$ or any ray zero to infinity.

dustin
  • 8,559
  • 1
    I think you wanted to say by $f(1) = 0\mapsto f(e^{2\pi i}) = 2\pi i$. Therefore, $f(z)$ is multivalued. – L.K. Mar 24 '17 at 10:38