Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
$a \notin \mathbb{Z}$, right? –  Cocopuffs Aug 24 '12 at 15:21
    
$a\in\mathbb C\setminus\mathbb Z$. I edited the question. –  Galoisfan Aug 24 '12 at 15:27
    
Which ones? The case $n$ nonzero integer, the case $a$, the case $0$, or all of them... –  Did Aug 24 '12 at 15:32
    
I don't understand all of them. –  Galoisfan Aug 24 '12 at 15:38
1  
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. –  Did Aug 24 '12 at 15:51
add comment

2 Answers

up vote 1 down vote accepted

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 $\frac1a$

share|improve this answer
add comment

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.


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


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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.