# Mittag-Leffler's theorem

In complex analysis, Mittag-Leffler's theorem concerns the existence of meromorphic functions with prescribed poles. Conversely, it can be used to express any meromorphic function as a sum of partial fractions. It is sister to the Weierstrass factorization theorem, which asserts existence of holomorphic functions with prescribed zeros. It is named after Gösta Mittag-Leffler.

## Theorem

Let ${\displaystyle D}$ be an open set in ${\displaystyle \mathbb {C} }$ and ${\displaystyle E\subset D}$ a closed discrete subset. For each ${\displaystyle a}$ in ${\displaystyle E}$, let ${\displaystyle p_{a}(z)}$ be a polynomial in ${\displaystyle 1/(z-a)}$. There is a meromorphic function ${\displaystyle f}$ on ${\displaystyle D}$ such that for each ${\displaystyle a\in E}$, the function ${\displaystyle f(z)-p_{a}(z)}$ has only a removable singularity at ${\displaystyle a}$. In particular, the principal part of ${\displaystyle f}$ at ${\displaystyle a}$ is ${\displaystyle p_{a}(z)}$.

One possible proof outline is as follows. Notice that if ${\displaystyle E}$ is finite, it suffices to take ${\displaystyle f(z)=\sum _{a\in E}p_{a}(z)}$. If ${\displaystyle E}$ is not finite, consider the finite sum ${\displaystyle S_{F}(z)=\sum _{a\in F}p_{a}(z)}$ where ${\displaystyle F}$ is a finite subset of ${\displaystyle E}$. While the ${\displaystyle S_{F}(z)}$ may not converge as F approaches E, one may subtract well-chosen rational functions with poles outside of D (provided by Runge's theorem) without changing the principal parts of the ${\displaystyle S_{F}(z)}$ and in such a way that convergence is guaranteed.

## Example

Suppose that we desire a meromorphic function with simple poles of residue 1 at all positive integers. With notation as above, letting

${\displaystyle p_{k}={\frac {1}{z-k}}}$

and ${\displaystyle E=\mathbb {Z} ^{+}}$, Mittag-Leffler's theorem asserts (non-constructively) the existence of a meromorphic function ${\displaystyle f}$ with principal part ${\displaystyle p_{k}(z)}$ at ${\displaystyle z=k}$ for each positive integer ${\displaystyle k}$. This ${\displaystyle f}$ has the desired properties. More constructively we can let

${\displaystyle f(z)=z\sum _{k=1}^{\infty }{\frac {1}{k(z-k)}}}$.

This series converges normally on ${\displaystyle \mathbb {C} }$ (as can be shown using the M-test) to a meromorphic function with the desired properties.

## Pole expansions of meromorphic functions

Here are some examples of pole expansions of meromorphic functions:

${\displaystyle {\frac {1}{\sin(z)}}=\sum _{n\in \mathbb {Z} }{\frac {(-1)^{n}}{z-n\pi }}={\frac {1}{z}}+2z\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{z^{2}-(n\,\pi )^{2}}}}$
${\displaystyle \cot(z)\equiv {\frac {\cos(z)}{\sin(z)}}=\sum _{n\in \mathbb {Z} }{\frac {1}{z-n\pi }}={\frac {1}{z}}+2z\sum _{k=1}^{\infty }{\frac {1}{z^{2}-(k\,\pi )^{2}}}}$
${\displaystyle {\frac {1}{\sin ^{2}(z)}}=\sum _{n\in \mathbb {Z} }{\frac {1}{(z-n\,\pi )^{2}}}}$
${\displaystyle {\frac {1}{z\sin(z)}}={\frac {1}{z^{2}}}+\sum _{n\neq 0}{\frac {(-1)^{n}}{\pi n(z-\pi n)}}={\frac {1}{z^{2}}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n\,\pi }}{\frac {2z}{z^{2}-(n\,\pi )^{2}}}}$