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I am really struggling with the concept of a "branch point". I understand that, for example, if we take the $\log$ function, by going around $2\pi$ we arrive at a different value, so therefore it is a multivalued function. However, surely this argument holds for all points in the complex plane, so I don't really understand how $z=0$ is the ONLY branch point.

Additionally, the course I am revising for needs no Riemann surfaces or knowledge of that area of mathematics, just what a branch point is and how to find it.

Thanks for any help.

iadvd
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user122316
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1 Answers1

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A branch point of a "multi-valued function" $f$ is a point $z$ with this property: there does not exist an open neighbourhood $U$ of $z$ on which $f$ has a single-valued branch.

In the case of $\log$, the only branch point is $0$: indeed, if $z \ne 0$, we could take $$U = \{ w \in \mathbb{C} : \lvert w - z \rvert < \lvert z \rvert \}$$ and define a single-valued branch of $\log$ on $U$. If you want to think in terms of paths, the point is that the value of $\log$ cannot jump if the path does not wind around $0$.

Zhen Lin
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    Thank you for your answer. I am still unsure as to how 0 is the only branch point, for example if we choose z=1 and go around the circle, we do not return to z=1, so therefore why is z=1 not a branch point? – user122316 May 14 '15 at 13:04
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    That is because the circle is too big and does in fact wind around $0$. – Zhen Lin May 14 '15 at 13:46
  • @ZhenLin Based on your definition of neighborhood $U$, since the circle will always include $z=0$, then aren't you saying that $log(0)$ is a single-valued branch? and if not, then what is a "single-valued branch"? – makansij Nov 21 '19 at 04:04
  • The definition in this 2015 answer disagrees with standard usage for algebraic functions. Under the usual definition, $\sqrt{1-\sqrt{z}}$ has a branch point at $z=1$, but there is a single-valued branch of $\sqrt{z}$ in a neighborhood of $z=1$ taking the value $-1$ at $z=1$, and hence a single-valued branch of $\sqrt{1-\sqrt{z}}$ in a neighborhood of $z=1$ taking the value $\sqrt{2}$ at $z=1$. Another way to say it: the Riemann surface of this function is ramified at some but not all of the points above $z=1$; that's enough to make $1$ a branch point. – Bjorn Poonen Jan 16 '21 at 05:17