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I was given a problem to prove a theorem by Mittag-Leffler about prescribing the items in the title, using Weierstrass's theorem about prescribed zeroes and Mittag-Leffler's theorem about prescribed poles and principal parts. I have worked something out and I would like to see if people agree and/or whether there are other approaches.

The statement to be proved: Let $U\subset\mathbb{C}$ be an open set and $\{a_{n}\}_{n=1}^{\infty}$ a sequence of distinct points in $U$ with no accumulation points in $U$. Let $\{p_{n}\}_{n=1}^{\infty}$ and $\{q_{n}\}_{n=1}^{\infty}$ be sequences of polynomials. There exists a meromorphic function $f$ on $U$ having no zeroes or poles outside the set $\{a_{n}:n\in\mathbb{N}\}$ and such that for each $n$, the difference $f(z)-p_{n}(\frac{1}{z-a_{n}})-q_{n}(z-a_{n})$ has (a removable singularity and) a zero at $a_{n}$ of order at least $1+\deg q_{n}$.

My attempt: By the Weierstrass theorem, there is a holomorphic function $g$ on $U$ having for each $n$ a zero of order $1+\deg q_{n}$ at $a_{n}$, and no other zeroes. By the Mittag-Leffler theorem, there is a meromorphic function $h$ having poles at each $a_{n}$ and for each $n$ the same principal part at $a_{n}$ as that of $\frac{p_{n}(1/(z-a_{n}))-q_{n}(z-a_{n})}{g(z)}$. (This quotient has poles at each $a_{n}$ by the choice of order of zeroes of $g$.) Let $f_{1}=gh$. Then $f_{1}$ has zeroes at some of the $a_{n}$'s and possibly some others, and has poles only at some of the $a_{n}$'s. Use the Weierstrass theorem to obtain holomorphic functions $a,b$ on $U$ such that $a$ has zeroes exactly at the $a_{n}$'s where $f_{1}$ has zeroes and $b$ has zeroes exactly where $f_{1}$ has poles. Then $f_{2}=\frac{a}{b}$ has zeroes and poles of the same orders as $f_{1}$ at the $a_{n}$'s and has no other zeroes or poles. Now we need to control the Laurent series coefficients. Locally at each $a_{n}$, we can define a holomorphic branch of $\log\frac{f_{1}}{f_{2}}$, and there is a holomorphic function $\phi$ on $U$ that agrees to suitably high order at each $a_{n}$ with $\log\frac{f_{1}}{f_{2}}$. Then $f=f_{2}e^{\phi}$ agrees with $f_{1}$ locally at each $a_{n}$ to that order so it has the desired principal part and finitely many terms of positive exponent, as well as the desired zeroes and poles.

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