theorem of Mittag-Leffler I want to find a meromorphic function that has poles exactly in the natural numbers and has the principal part $\frac{1}{z-n}.$
Therefore I used the theorem of Mittag-Leffler
and I received $\sum_{n=1}^{\infty}(\frac{1}{z-n}+\frac{1}{n}).$ Is this correct?
And unfortunately I don't quite understand how exactly I can read the principle part..
Thanks for your help!!
 A: Not sure what you mean by "principle part". I think you mean the Laurent series, which is going to be different for each singularity of a function.
I think you are mixing up a Laurent series vs a rational series. A Laurent series is of the form:
$$f(z) = \sum_{k=-\infty}^{\infty} a_k(z-z_0)^k $$
Notice how the series is centered at $z_0$, which is a singularity to the meromorphic function. A rational series is
$$ f(z) = \sum_{k=0}^\infty \frac{a_k}{z - b_k}$$
Notice that there is no "center". This is not a Laurent series, does not have an annulus of convergence and is instead governed by different theorems. However, you can use any sequence of convergent analytic functions to discover properties about the Laurent series.
For example, if I wanted coefficient $a_{-1}$ in the Laurent series centered at $n$, for the function defined by the rational series
$$f(z) = \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n}$$
Then all I have to do is integrate around a small compact contour $C$ that is centered at $n$, canceling out those terms that integrate to zero:
$$\begin{align} \int_C \sum_{k=-\infty}^{\infty} a_k (z-n)^k &= \int_C \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n} \\ a_{-1}\cdot 2\pi i &= \int_C \frac{1}{z-n}  \\ a_{-1}\cdot 2\pi i &= 2\pi i \\ a_{-1} &= 1 \end{align}$$
So now we know that the Laurent series centered at $n$ has a principle coefficient of $1$. Again, note there is a different Laurent series for each singularity.
