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I am looking for ideas on proving the Eulerian expansion of the cotangent using residue calculation: $$\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^{\infty}\left(\frac{1}{z+n}+\frac{1}{z-n}\right), \ z\in\mathbb{C}\backslash\mathbb{Z}.$$

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    $\begingroup$ Although your question is about the cotangent function and complex analysis, the tag “cotangent-complex” is actually not suitable since that means something else entirely: see en.m.wikipedia.org/wiki/Cotangent_complex. $\endgroup$
    – KConrad
    Commented May 23, 2023 at 15:56

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The residue approach to the partial fraction expansion of $\cot(z)$ is explained in Freitag & Busam's "Complex Analysis", Prop. III.7.13, and probably in other books as well.

Here is a more general result regarding partial fraction expansions:

Proposition: Let $f \in \mathcal{M}(\mathbb{C})$ and let $(\gamma_n)_{n \in \mathbb{N}}$ be a family of shrinking rectifiable loops that avoid the poles of $f$ and whose interiors exhaust $\mathbb{C}$, i.e. $\gamma_n$ is in the interior of $\gamma_{n+1}$. If $$ \lim_{n \to \infty} \int_{\gamma_n} \frac{|f(z)| |dz|}{|z|} = 0, $$ then $$ f(z) = \sum_{n=1}^\infty \sum_{p_k \in A_n} P_k\bigg(\frac{1}{z-p_k}\bigg), $$ where $A_n :=$ interior of $\gamma_n$ $\setminus$ closure of the interior of $\gamma_{n-1}$ is the $n$-th open "annulus", $p_k$ are the poles of $f$ lying in the respective open annulus and $P_k$ are the corresponding principal parts.

Proof: Apply the residue theorem for each $\gamma_n$ and use the vanishing condition to ensure (compact) convergence. $\square$

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