$\textbf{Problem:}$
Find a meromorphic function that has simple poles in $ \, n = 1,2,3, \dots \,$ and in $\, m =0, i, 2i,3i, \dots \,$ (the residue at these poles is not specified so we set it to 1).
$\textbf{What i know:} $ Firstly let's define the sequence of zeroes as $s_n := \frac{n+1}{2} \,$ for $n$ odd and $s_n:= \frac{in}{2}\,$ for $n$ even. Such that the sequence looks like this: $\quad 1 ,i,2,2i, 3 ,\dots \quad $
Now i want to use Mittag-Leffler's theorem (here: https://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem). Using the notation in the link, the function i'm looking for should be similar to this:
$$
f(z) = \sum^{}_{n\geq0} p_n(z) \qquad\text{where} \qquad p_n(z) = \frac {1}{z-s_n}
$$
I think the series above diverges, so usually what you do is:
write $p_n(z)$ as a power series and remove the part that diverges from the series above, i should come up with something like this: (not this particular coefficients but in general a polynomial in z)
$$
f(z) = \sum^{}_{n\geq0}p_n(z) \, – \, \big(1 \, + \, \frac{z}{s_n} \, + \, \frac{z^2}{s_n^2} \, \,+\, \dots\big)
$$
where in the second part of the substraction i should keep deleting terms from the series until i get something that is $\,O(\frac{1}{n^2})\,$ or in general a term that converges.
However i don't know how to proceed in practice, i would need help on how to write the power series, or in general on how to solve this problem using Mittag-Leffler's theorem.
Best Answer
You can simplify the problem by looking for a function $g$ with simple poles at the positive integers first. Then $$ f(z) = \frac 1z + g(z) + g(-iz) $$ has the required properties.
$g$ can be constructed as a series of the form $$ g(z) = \sum_{n=1}^\infty \frac{1}{z-n} - T_n(z) $$ where $T_n(z)$ is a Taylor polynomial of $\frac{1}{z-n}$ at $z=0$ of sufficiently high degree to make the series converge (uniformly on compact sets). Now $$ \frac{1}{z-n} = -\frac 1n - \frac{z}{n^2} - \frac{z^2}{n^3} -\ldots $$ and the first attempt of subtracting just the constant term (i.e. the Taylor polynomial of degree zero) works in our case $$ g(z) = \sum_{n=1}^\infty \frac{1}{z-n} +\frac 1n = \sum_{n=1}^\infty \frac{z}{(z-n)n} $$ because the denominator grows quadratic in $n$.