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

[fauboca]

If f is meromorphic on U with only a finite number of poles, then f=\frac{g}{h} 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 z_0 is such a point, then there exist m in integers such that (z-z_0)^mf(z) is holomorphic in a neighborhood of z_0.

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

S0 the trouble is showing that f is the quotient of two holomorphic functions.

 

[Dick]

How would you show it if there is only one pole?

 

[fauboca]

How would you show it if there is only one pole?

If the pole is at z_0, then

f=\frac{g}{(z-z_0)}

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.

fh=g\Rightarrow f=\frac{g}{h}

 

[Dick]

If the pole is at z_0, then

f=\frac{g}{(z-z_0)}

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.

fh=g\Rightarrow f=\frac{g}{h}

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