when all singularities are poles

In the parent entry (http://planetmath.org/ZerosAndPolesOfRationalFunction) we see that a rational function has as its only singularities a finite set of poles. It is also valid the converse

Theorem. Any single-valued analytic function, which has in the whole closed complex plane no other singularities than poles, is a rational function.

Proof. Suppose that  $z\mapsto w(z)$  is such an analytic function. The number of the poles of $w$ must be finite, since otherwise the set of the poles would have in the closed complex plane an accumulation point which is neither a point of regularity (http://planetmath.org/Holomorphic) nor a pole. Let $b_1,b_2,\mathrm{\dots },b_k$ and possibly $\mathrm{\infty }$ be the poles of the function $w$.

For every  $i=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots },k$,  the function has at the pole $b_i$ with the order $n_i$, the Laurent expansion of the form

$w(z)={\displaystyle \frac{c_{-n_i}}{{(z-b_i)}^{n_i}}}+{\displaystyle \frac{c_{-n_i+1}}{{(z-b_i)}^{n_i-1}}}+\mathrm{\dots }+c_0+c_1(z-b_i)+\mathrm{\dots }$(1)

This is in in the greatest open disc containing no other poles. We write (1) as

$w(z)=F_{n_i}\left({\displaystyle \frac{1}{z-b_i}}\right)+P(z-b_i),$(2)

where the first addend is the principal part of (1), i.e. consists of the terms of (1) which become infinite in  $z=b_i$.

If we think a circle having center in the origin and containing all the finite poles $b_i$ (an annulus ), then $w(z)$ has outside it the Laurent series expansion

$$w(z)=d_m{z}^m+d_{m-1}{z}^{m-1}+\mathrm{\dots }+d_0+\frac{d_{-1}}{z}+\mathrm{\dots },$$

which we write, corresponding to (2), as

$w(z)=G_m(z)+Q\left({\displaystyle \frac{1}{z}}\right),$(3)

where $G_m(z)$ is a polynomial of $z$ and $Q\left(\frac{1}{z}\right)$ a power series in $\frac{1}{z}$. Then the equation

$$R(z):=\sum _{i=1}^k{F}_{n_i}\left(\frac{1}{z-b_i}\right)+G_m(z)$$

defines a rational function having the same poles as $w$. Therefore the function defined by

$$f(z):=w(z)-R(z)$$

is analytic (http://planetmath.org/Analytic) everywhere except possibly at the points  $z=b_i$  and  $z=\mathrm{\infty }$.  If we write

$$f(z)=\left[w(z)-F_{n_i}\left(\frac{1}{z-b_i}\right)\right]-\sum _{j\ne i}{F}_{n_j}\left(\frac{1}{z-b_j}\right)-G_m(z),$$

we see that $f(z)$ is bounded in a neighbourhood of the point $b_i$ and is analytic also in this point ($i=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots },k$). But then again, the

$$f(z)=\left[w(z)-G_m(z)\right]-\sum _{j=1}^k{F}_{n_j}\left(\frac{1}{z-b_j}\right)$$

shows that $f$ is analytic in the infinity (http://planetmath.org/RiemannSphere), too. Thus $f$ is analytic in the whole closed complex plane. By Liouville’s theorem, $f$ is a constant function.  We conclude that  $R(z)+f(z)=w(z)$  is a rational function. Q.E.D.

The theorem implies, that if a meromorphic function is regular at infinity or has there a pole, then it is a rational function.