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Multiplicity of a pole

  1. May 28, 2008 #1
    In complex analysis, what is understood by the multiplicity of a pole?

    thank you
  2. jcsd
  3. May 28, 2008 #2
    For [tex]f(z)=\frac{1}{(z-z_0)^n}[/tex], the pole at [tex]z=z_0[/tex] has multiplicity n
  4. May 28, 2008 #3


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    i.e. the order of the pole of f at z, equals the multiplicity of the zero of 1/f, at z.
  5. May 28, 2008 #4
    May this be extended beyond polynomials?
  6. May 28, 2008 #5
  7. May 28, 2008 #6


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    it makes sense for the quotient of two holomorphic power series, yielding what is called a laurent series, i.e., a series with at most a finite number of terms of negative powers.

    on any open set, the fraction field of the holomorphic functions form what is called the field of meromorphic functions on that set. those have at worst poles as non holomorphic points.

    e^(1/z) has a worse than pole point at z = 0. isolated non holomorphic points are called (isolated) singularities. the simplest actual singularities of functions defined by power series, possibly infinite in both directions, i.e. summed over all integer powers of z, are the poles.
  8. May 28, 2008 #7
    I.e. the multiplicity is the power of the term with the largest negative power in the laurent series of the function?

    Does this also mean that an isolated/(essential?) singularity is a pole with infinite multiplicity?
  9. May 28, 2008 #8


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    More generally, a function f(z) has a pole at z= a of multiplicity n if and only if
    [itex]\lim_{z\rightarrow a} (z-a)^nf(z)[/itex] exists but [itex]\lim_{z\rightarrow a} (z-a)^{n-1}f(z)[/itex] does not.
  10. May 28, 2008 #9
    That definition put everything in place, thanks!
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