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

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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.
 

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Dick
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How would you show it if there is only one pole?
 
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How would you show it if there is only one pole?
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]
 
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Dick
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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]
You want h to have 'zeroes', not 'singularities'.
 

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