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

One possible proof outline is as follows. If $E$  is finite, it suffices to take $f(z)=\sum _{a\in E}p_{a}(z)$ . If $E$  is not finite, consider the finite sum $S_{F}(z)=\sum _{a\in F}p_{a}(z)$  where $F$  is a finite subset of $E$ . While the $S_{F}(z)$  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 $S_{F}(z)$  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

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

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

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

This series converges normally on $\mathbb {C}$  (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:

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