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Singularities: Complicated function

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all the singularities of [tex]f(z) = \frac{{{e^{\frac{1}{{z - 1}}}}}}{{{e^{\frac{1}{z}}} - 1}}[/tex] in the extended complex field, classify them and find Res(f, 0) and Res(f, infinity)


    2. Relevant equations

    Res(f, z0) = a-1 in the Laurent series around that z0

    [tex]{e^z} = \sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{n!}}} [/tex]

    3. The attempt at a solution

    It's kind of a difficult function to find singularities from.

    First I tried to look at the numerator. I found that 1 is an essential singularity, because the Laurent series of [tex]{{e^{\frac{1}{{z - 1}}}}}[/tex] around 1 is [tex]\sum\limits_{n = o}^\infty {\frac{1}{{n!{{(z - 1)}^n}}}} [/tex], which is a series formed by negative powers of (z-1) only. So the series has infinite negative powers of (z-1)

    Then I tried to look at the denominator. I know zero has to be a singularity of some sort, but I can't find a way to classify it or to justify that.

    Finally, for the infinity, I make the variable change w = 1/z, then I evaluate [tex]\frac{{{e^{\frac{w}{{1 - w}}}}}}{{{e^w} - 1}}[/tex] when w=0. I find that zero is a simple pole of f(1/w) (because 1/f(1/w) has a simple zero in w=0), then infinity a simple pole of f(z).

    How do I find what kind of singularity is z=0?

    Thanks
     
  2. jcsd
  3. Oct 26, 2009 #2
    Anyone?
     
  4. Oct 26, 2009 #3
    Nobody?
     
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