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In the parent entry 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 nor a pole. Let $b_{1},\,b_{2},\,\ldots,\,b_{k}$ and possibly $\infty$ be the poles of the function^{} $w$.

For every $i=1,\,2,\,\ldots,\,k$, the function has at the pole $b_{i}$ with the order $n_{i}$, the Laurent expansion of the form

$\displaystyle w(z)=\frac{c_{{-n_{i}}}}{(z-b_{i})^{{n_{i}}}}+\frac{c_{{-n_{i}+1% }}}{(z-b_{i})^{{n_{i}-1}}}+\ldots+c_{0}+c_{1}(z-b_{i})+\ldots$ | (1) |

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

$\displaystyle w(z)=F_{{n_{i}}}\!\left(\frac{1}{z-b_{i}}\right)+P(z-b_{i}),$ | (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^{}
$\varrho<|z|<\infty$), then $w(z)$ has outside it the Laurent series expansion

$w(z)=d_{m}z^{m}+d_{{m-1}}z^{{m-1}}+\ldots+d_{0}+\frac{d_{{-1}}}{z}+\ldots,$ |

which we write, corresponding to (2), as

$\displaystyle w(z)=G_{m}(z)+Q\!\left(\frac{1}{z}\right)\!,$ | (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 everywhere except possibly at the points $z=b_{i}$ and $z=\infty$. If we write

$f(z)=\left[w(z)-F_{{n_{i}}}\!\left(\frac{1}{z-b_{i}}\right)\right]-\sum_{{j% \neq i}}F_{{n_{j}}}\!\left(\frac{1}{z-b_{j}}\right)-G_{m}(z),$ |

we see that $f(z)$ is bounded^{} in a neighbourhood of the point $b_{i}$ and is analytic also in this point ($i=1,\,2,\,\ldots,\,k$). But then again, the presentation

$f(z)=\left[w(z)-G_{m}(z)\right]-\sum_{{j=1}}^{k}F_{{n_{j}}}\!\left(\frac{1}{z-% b_{j}}\right)$ |

shows that $f$ is analytic in the infinity, 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.

# References

- 1 R. Nevanlinna & V. Paatero: Funktioteoria. Kustannusosakeyhtiö Otava, Helsinki (1963).

## Mathematics Subject Classification

30D10*no label found*30C15

*no label found*30A99

*no label found*

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