# A question about branch cut

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.

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This may not exactly answer your question, but my answer here may be helpful. It doesn't talk about Riemann surfaces, but it does go in depth about the existence of logarithm of holomorphic functions. –  J. Loreaux Oct 8 '12 at 2:42
For this sort of question, it's a good idea to state concretely what it is about the Wikpedia article on the subject that you find lacking or hard to understand; then we can fill that in specifically instead of writing an entire article from scratch. –  joriki Oct 8 '12 at 5:54
Not exactly about Riemann surfaces, but if you are interested in branch cuts and branch points in general, then these notes 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. –  EuYu Oct 8 '12 at 5:57

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.