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Theorem III of Townshends Complex Analysis has a proof that rational functions have no essential singularities.

Just wondering, are there any particularly elegant proofs that a rational function has no essential singularities?

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They are quotients of polynomials, which are meromorphic functions on the sphere—therefore rational functions are also meromorphic on the sphere. This reduces to checking that polynomials have a pole at $\infty$, which is easy; for example, they grow polynomially :) – Mariano Suárez-Alvarez May 6 '12 at 7:44
@Mariano: Assuming they have a singularity at $\infty$, i.e. are not constant. :) – anon May 6 '12 at 7:52
Yeah, that :D (Well... constant polynomials have at most a pole at infinity :) ) – Mariano Suárez-Alvarez May 6 '12 at 7:58
Do you pdf file of this book? – Sushil yesterday

1 Answer 1

up vote 4 down vote accepted

Since any rational function is of the form $f(z)=\frac{p(z)}{q(z)}$, where $p,q$ are polynomials, the only singularities of $f(z)$ are the zeroes of $q(z)$ and $\infty$. $q(z)$ is a polynomial, hence all it's zeros are of finite order - so they are poles of that order (assuming $f$ is reduced, i.e. $p,q$ have no common zeroes). Depending on the degrees of $p,q$ it is easy to check that $\lim_{z\to\infty}f(z)$ is either $0$, $\infty$ or the quotient of the leading coefficients of $p,q$. In any case it exists, so $\infty$ can not be essential singularity.

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