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

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.

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Dick
Homework Helper

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

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.

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

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Dick
Homework Helper

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'.