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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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: |
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