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Riemann surfaces

by unchained1978
Tags: riemann, surfaces
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unchained1978
#1
Jan1-13, 06:36 PM
P: 99
I don't mean to sound ignorant, but when reading up on complex analysis in the broad sense, I don't really see the point of introducing Riemann surfaces. It's a way of making multivalued functions single valued, but so what? I don't see the utility of such an idea, which isn't to argue there is none, I just don't understand it that well. Can someone explain why you 'need' Riemann surfaces or how they actually help, or are they just an alternative way of looking at complex functions?
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mathwonk
#2
Jan1-13, 07:27 PM
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i wrote this answer up and posted it somewhere on here. i will try to find it for you.


see post #3 here:

http://www.physicsforums.com/showthread.php?t=553454
jackmell
#3
Jan4-13, 07:33 AM
P: 1,666
Quote Quote by unchained1978 View Post
I don't mean to sound ignorant, but when reading up on complex analysis in the broad sense, I don't really see the point of introducing Riemann surfaces. It's a way of making multivalued functions single valued, but so what? I don't see the utility of such an idea, which isn't to argue there is none, I just don't understand it that well. Can someone explain why you 'need' Riemann surfaces or how they actually help, or are they just an alternative way of looking at complex functions?
The entire foundation of Complex Analysis of analytic single-valued functions can be applied to multi-valued functions when they are mapped to Riemann surfaces. That's because they become single-valued analytic functions of the coordinates on this new coordinate system, the Riemann surface. Take a meromorphic function and the expression:

[tex]\oint f(z)dz=2\pi i \sum r_i[/tex]

we know that. Now the remarkable fact, is that the Residue Theorem, Cauchy's Integral formula, and theorem, Gauss's Mean Value Theorem, the Argument Theorem, Laurent's expansion Theorem, and the rest can also be applied to multi-valued functions when they are mapped to their Riemann surfaces like for example:

[tex]\oint \sqrt[5]{f(z)}dz=2\pi i\sum q_i[/tex]

except the path is not over the z-plane but rather the Riemann surface of the function and the residues [itex]q_i[/itex] are the residues of the assoicated Laurent-Puiseux expansions. Same dif for the rest of the theorems in Complex Analysis.


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