# 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. 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:

${\displaystyle \tan(z)=\sum \limits _{n=0}^{\infty }{\dfrac {8z}{(2n+1)^{2}\pi ^{2}-4z^{2}}}}$
${\displaystyle \csc(z)=\sum _{n\in \mathbb {Z} }{\frac {(-1)^{n}}{z-n\pi }}={\frac {1}{z}}+2z\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{z^{2}-(n\,\pi )^{2}}}}$
${\displaystyle \sec(z)\equiv -\csc \left(z-{\frac {\pi }{2}}\right)=\sum _{n\in \mathbb {Z} }{\frac {(-1)^{n-1}}{z-\left(n+{\frac {1}{2}}\right)\pi }}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n+1)\pi }{(n+{\frac {1}{2}})^{2}\pi ^{2}-z^{2}}}}$
${\displaystyle \cot(z)\equiv {\frac {\cos(z)}{\sin(z)}}=\sum _{n\in \mathbb {Z} }{\frac {1}{z-n\pi }}={\frac {1}{z}}+2z\sum _{k=1}^{\infty }{\frac {1}{z^{2}-(k\,\pi )^{2}}}}$
${\displaystyle \csc ^{2}(z)=\sum _{n\in \mathbb {Z} }{\frac {1}{(z-n\,\pi )^{2}}}}$
${\displaystyle \sec ^{2}(z)={\dfrac {d}{dz}}\tan(z)=\sum \limits _{n=0}^{\infty }{\dfrac {8((2n+1)^{2}\pi ^{2}+4z^{2})}{((2n+1)^{2}\pi ^{2}-4z^{2})^{2}}}}$
${\displaystyle {\frac {1}{z\sin(z)}}={\frac {1}{z^{2}}}+\sum _{n\neq 0}{\frac {(-1)^{n}}{\pi n(z-\pi n)}}={\frac {1}{z^{2}}}+\sum _{n=1}^{\infty }{(-1)^{n}}{\frac {2}{z^{2}-(n\,\pi )^{2}}}}$