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I Residue Theorem

  1. Jun 6, 2017 #1
    Hello! Why do the singularities in the Residue Theorem must be isolated? If we have let's say a disk around ##z_0##, ##D_{[z_0,R]}## where all the points are singularities for a function ##f:G \to C## with the disk in region G, but f is holomorphic in ##G-D_{[z_0,R]}##, we can still write f as a Laurent series in ##G-D_{[z_0,R]}## and thus we still have ##\int_{\gamma}f(z)dz = 2\pi i c_{-1}##, with ##\gamma## a picewise, smooth, closed path in ##G-D_{[z_0,R]}##. I know that ##c_{-1}## is called residue only if the singularity is isolated, but letting aside the nomenclature, why can't we use the formula for non-isolated singularities?
     
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  3. Jun 6, 2017 #2

    FactChecker

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    If the singularity is not isolated, the Laurent series will not converge in any region around the center of expansion.
     
  4. Jun 6, 2017 #3
    But isn't the Laurent series defined in an annulus of convergence? Which means that the inner disk doesn't matter (f can have or not infinitely many singularities in there) but only the region between the inner and outer disk?
     
  5. Jun 6, 2017 #4

    mathwonk

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    Indeed the residue theorem is true and sometimes stated in the more general case you mention, e.g. in the book of Henri Cartan, Prop. 2.1, page 91. As he says there however, the interest of the isolated case is that there one can often actually calculate the residue, at least for poles.
     
  6. Jun 6, 2017 #5
    Thank you for you reply. So it is true even for the case I mentioned, but practically it is useful for isolated singularities, right?
     
  7. Jun 6, 2017 #6

    mathwonk

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    right.
     
  8. Jun 6, 2017 #7

    mathwonk

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    I am not an expert, but I seem to recall, to construct a laurent series converging in the annulus r < |z| < s, you take one power series converging in the disc |z| < s, and another series with no constant term converging in the disc |z| < 1/r. Then you replace z with 1/z everywhere in the second series, obtaining one that converges outside the circle |z| = r. adding this "negative" power series to the first one gives a laurent series converging in the intersection of the two "discs" |z| < s, and |z| > r.
     
  9. Jun 6, 2017 #8

    FactChecker

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    Sorry. Right. I stand corrected.
     
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