I am trying to understand the derivation of the partial fraction decomposition of the cotangent. Therefore I have some notes but unfortunately there is a little detail I do not understand.

We are considering

\begin{align} f: \mathbb{C}\setminus \mathbb{Z} \to \mathbb{C}, \quad f(z):= \pi \cot(\pi z) = \frac{\pi \cos(\pi z)}{\sin(\pi z)} \end{align} and we are trying to prove that

\begin{align} \frac{1}{z} + \sum_{n=1}^{\infty} \left(\frac{1}{z+n} + \frac{1}{z-n}\right) = \pi \cot(\pi z). \end{align}

With the help of the following statement

Theorem: Let $D \subset \mathbb{C}$ open, $z_0 \in D$ and $g,h: D \to \mathbb{C}$ holomprohic, where $g(z_0) \neq 0$, $h(z_0) = 0$ and $h'(z_0) \neq 0$.Then there exists a neighbourhood $U$ of $z_0$ such that

\begin{align} f: U \setminus \{z_0\} \to \mathbb{C}, \quad f(z):= \frac{g(z)}{h(z)} \end{align} is holomorphic and has a simple pole at $z_0$ with \begin{align} \operatorname{Res}_{z_0}f = \frac{g(z_0)}{h'(z_0)}. \end{align}

So far so good. Now applying this Theorem on our function $f$ yields that, for $n_0 \in \mathbb{Z}$, we can write

\begin{align} f(z) = \frac{1}{z-n_0} + p(z) \quad \forall z \in U\setminus \{n_0\} \end{align}

where the first term is the principle part (with residue $1$ (follows from Theorem)) of a Laurent series of $f$ - the Laurent series we are interested in, since we want to derive the partial fraction decomposition. $p(z)$ is the regular part of the laurent series. We only know that $p(z)$ is holomorphic.

Based on these oberservations the notes "guess" that the partical defraction decomposition of the cotangent could look like

\begin{align} f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} \left(\frac{1}{z+n} + \frac{1}{z-n}\right). \end{align}

From now on they proof that it actually is the partical defraction decomposition by showing that

\begin{align} \frac{1}{z} + \sum_{n=1}^{\infty} \left(\frac{1}{z+n} + \frac{1}{z-n}\right) \end{align}

is holomorphic and - in the second part - is identical to $\pi \cot(\pi z)$. That second part is where my question can be found.

In order to show that

\begin{align} \frac{1}{z} + \sum_{n=1}^{\infty} \left(\frac{1}{z+n} + \frac{1}{z-n}\right) \end{align}

and $\pi \cot(\pi z)$ are identical we consider the help function

\begin{align} h(z) := \frac{1}{z} + \sum_{n=1}^{\infty} \left(\frac{1}{z+n} + \frac{1}{z-n}\right) - \pi \cot(\pi z). \end{align}

We obviously try showing that $h = 0$. And finally to my question:

Question: The note states that $h$ can be extended to an entire fuction since "the regular parts of the laurent series cancel each other out".

I don't get that. Shouldn't it be the principal parts?


Yes, it should be the principal parts.


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