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Every meromorphic function on $C$ is a quotient of entire functions

Statement

Every meromorphic function on $C$ is a quotient of entire functions.

Facts & Assumptions

Given: A meromorphic function $f$ on $C$ .

[F1]

A meromorphic function on a plane domain is holomorphic off a discrete pole set, and each pole is an isolated pole in the usual sense (Meromorphic functions on a plane domain).

[F2]

The Weierstrass product theorem constructs an entire function with any prescribed discrete zero divisor on $C$ (Weierstrass product theorem on the complex plane).

[F3]

A bounded punctured-neighbourhood singularity is removable (Characterizations of removable singularities).

Proof

technique · direct

1.1

Let $(p_{n})$ be the poles of $f$ , listed with multiplicity equal to the order of the pole. By [F2], there is an entire function $q$ whose zeros are exactly the points $p_{n}$ with those multiplicities and with no other zeros.

F1F2givenconstruct

2.1

On $C ∖ {p_{n}}$ define $g = f q$ . Near a pole $p_{n}$ of order $m$ , the zero of $q$ has the same order $m$ , so the product $g$ is locally bounded on the punctured neighbourhood of $p_{n}$ . By [F3], each such singularity is removable, hence $g$ extends to an entire function on $C$ .

F1F3step 1.1algebra

3.1

Away from the poles, $f = g / q$ by construction, and both sides are meromorphic with the same removed singularities at the poles. Therefore $f$ is the quotient of the two entire functions $g$ and $q$ .

step 1.1step 2.1algebra∎

Depends on

Used by

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 The Weierstrass product theorem (standard reference, not scraped)

Every meromorphic function on CC is a quotient of entire functions