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If f is meromorphic on U with only a finite number of poles, then

  1. Feb 25, 2012 #1
    If f is meromorphic on U with only a finite number of poles, then [itex]f=\frac{g}{h}[/itex] where g and h are analytic on U.

    We say f is meromorphic, then f is defined on U except at discrete set of points S which are poles. If [itex]z_0[/itex] is such a point, then there exist m in integers such that [itex](z-z_0)^mf(z)[/itex] is holomorphic in a neighborhood of [itex]z_0[/itex].

    A pole is [itex]\lim_{z\to a}|f(z)| =\infty[/itex].

    S0 the trouble is showing that f is the quotient of two holomorphic functions.
     
  2. jcsd
  3. Feb 25, 2012 #2

    Dick

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    Re: meromorphic

    How would you show it if there is only one pole?
     
  4. Feb 25, 2012 #3
    Re: meromorphic

    If the pole is at z_0, then

    [tex]f=\frac{g}{(z-z_0)}[/tex]

    Let S be the set of the poles of f in U and let h be analytic in U with singularities at the points in S. Let the order of the singularity be the order of the pole in f. Then fh has only removable singularities in U.

    [tex]fh=g\Rightarrow f=\frac{g}{h}[/tex]
     
    Last edited: Feb 25, 2012
  5. Feb 25, 2012 #4

    Dick

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    Re: meromorphic

    You want h to have 'zeroes', not 'singularities'.
     
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