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Riemann surfaces 
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#1
Jan113, 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?



#2
Jan113, 07:27 PM

Sci Advisor
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P: 9,488

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 


#3
Jan413, 07:33 AM

P: 1,666

[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 multivalued 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 zplane but rather the Riemann surface of the function and the residues [itex]q_i[/itex] are the residues of the assoicated LaurentPuiseux expansions. Same dif for the rest of the theorems in Complex Analysis. 


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