When we want to find a good domain such that the logarithm $f(z)=\log(z)$,we can let $\mathbb{C}\backslash\{z:z\le 0\}$ to be the domain of $f$ so that the logarithm is continuous. However, how can we find such domain if $f(z)=\log[(z-a_1)(z-a_2)...(z-a_n)]$ where $a_i$ are complex numbers? I have asked this question to my professor and he told my that it is related to something called Riemann surface. Can someone give me some reference or material so that I can know what he is talking about? Thank you.
A question about branch cut
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0Not exactly about Riemann surfaces, but if you are interested in branch cuts and branch points in general, then these [notes](http://scipp.ucsc.edu/~haber/ph116A/branch.pdf) may be an interesting reference. In particular, page $11$ of the notes covers a case similar to this problem, specifically $\log[(z+1)(z-1)]$. In general, introduction to Riemann surfaces tend to be covered by many standard Complex Analysis texts when discussing branch cuts. The subject itself is vast and can take some getting used to if you're just beginning complex analysis. – 2012-10-08
1 Answers
If you move around the unit circle $z = e^{2 \pi i t}$ then $\log z = 2 \pi i t$. This is a paradox since $z(t)$ is periodic while $\log z(t)$ is linear.
The resolution is that the plane as been "punctured" at one point $z = 0$ and now the surface you get is topologically a cylinder, where we identify $z \sim z + 2 \pi i$.
To take the log of a polynomial, $ f(z)=\log(z-a_1)(z-a_2)...(z-a_n) = \log(z-a_1) + \dots + \log(z-a_n)$ we add several punctures. We can add branch cuts from $z = z_n$ to $z = \infty$ leaving a surface which is simply-connected and has a "well-defined" logarithm.
These poles that we have introduced, are somewhat like "electric charges"... so we could say these charges alter the topology of our surface by making them not simply connected
For more discussion on branch cuts: