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Let be $a\in\mathbb C\setminus\mathbb Z$ a fixed complex number, and define the following function:

$$f(z)=\frac{a\pi\cot\pi z}{z(a-z)}$$ It has simple poles in $z\in\mathbb Z\setminus\{0\}$ and in $z=a$, and a double pole in $z=0$. my textbook says that it is straightforward that:

$\operatorname{Res}(f,n)=\frac{a}{n(a-n)}\;\;$ for $n\in\mathbb Z\setminus\{0\}$

$\operatorname{Res}(f,a)=-\pi\cot\pi a$

$\operatorname{Res}(f,0)=\frac{1}{a}$

but I don't understand why these fact are true.

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    $\begingroup$ $a \notin \mathbb{Z}$, right? $\endgroup$ – Cocopuffs Aug 24 '12 at 15:21
  • $\begingroup$ $a\in\mathbb C\setminus\mathbb Z$. I edited the question. $\endgroup$ – Dubious Aug 24 '12 at 15:27
  • $\begingroup$ Which ones? The case $n$ nonzero integer, the case $a$, the case $0$, or all of them... $\endgroup$ – Did Aug 24 '12 at 15:32
  • $\begingroup$ I don't understand all of them. $\endgroup$ – Dubious Aug 24 '12 at 15:38
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    $\begingroup$ If you know the definition of a residue at a pole, I find quite surprising that the values at $n\ne0$ and at $a$ escape you. $\endgroup$ – Did Aug 24 '12 at 15:51
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For simple poles, it's easy: $Res(f,n) = \lim_{z\to n} (z-n)f(z) = \frac{a}{n(a-n)}$ for $n \in \mathbb Z$ Because $\lim_{z\to n} (z-n) \cot(\pi z) = \frac{1}{\pi}$.

We also have: $Res(f,a) = \lim_{z\to a} (z-a)f(z) = -\pi \cot(\pi a)$

For the double pole in $b$, you can use the formula:

$Res(f,b) = \lim_{z\to b} \frac{d}{dz} \left( (z-b)^2f(z) \right)$

$\frac{d}{dz}\left( (z-b)^2f(z) \right) = \frac{-\pi\,a\,z\,{\csc\left( z\right) }^{2}}{a-z}+\frac{\pi\,a\,z\,\cot\left( z\right) }{{\left( a-z\right) }^{2}}+\frac{\pi\,a\,\cot\left( z\right) }{a-z}$

So the limit seems to be $\frac{\pi}{a}$ and not $\frac1a$

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Note that the residue of $\pi \cot (\pi z) g(z)$ at $z=n, n= 0, \pm1, \pm 2, \cdots$ is

$$\operatorname*{Res}_{z=n}\,\pi \cot (\pi z) g(z) = \lim_{z \to n} (z-n)\pi \cot (\pi z) g(z)= \lim_{z \to n} \pi \left(\frac{z-n}{\sin (\pi z)}\right) \cos (\pi z) g(z)=g(n)$$

It follows that

$$\operatorname*{Res}_{z=n}\,f(z)\,=\frac{a}{n(a-n)}$$

The reason that I used a general $g(x)$ function is because this interesting residue formula comes in handy for series evaluations later on.


$$\operatorname*{Res}_{z=a}\, \frac{a\pi\cot(\pi z)}{z(a-z)}=\lim_{z \to a} (z-a)\frac{a\pi\cot(\pi z)}{z(a-z)}=-\lim_{z \to a}\frac{a\pi\cot(\pi z)}{z}=-\pi \cot (\pi a)$$


$$\operatorname*{Res}_{z=0}\,f(z)=\lim_{z \to 0} (z-0)\frac{a\pi\cot(\pi z)}{z(a-z)}=\lim_{z \to 0} \frac{a\pi\cot(\pi z)}{a-z}=\frac{1}{a}$$

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