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.

What is

$$\Large{\mathrm{res}_{e^{\frac{i\pi}n}}\left(\frac1{z^n+1}\right)?}$$

My result doesn't agree with what WolframAlpha says.

I calculate it this way. We have

$$\Large{z^n+1=(z-e^{\frac{i\pi}n})\big(z-e^{\frac{3i\pi}n})\cdots(z-e^{\frac{(2n-1)i\pi}n})}$$

so, for $$\Large y=z-e^{\frac{i\pi}n},$$ we have

$$\Large{(z^n+1)^{-1}=y^{-1}\ (y-e^{\frac{i\pi}n}(1-e^{\frac{2i\pi}n}))^{-1}\ \cdots\ (y-e^{\frac{i\pi}n}(1-e^{\frac{(2n-2)i\pi}n}))^{-1}}.$$

I expand each of the factors around $y=0$, and obtain that each of them except the first has $$\Large e^{\frac{i\pi}n}(1-e^{\frac{2ki\pi}n})$$ as the constant coefficient of the series expansion. So the coefficient at $y^{-1}$ in the expansion of the whole product is

$$\Large 1\cdot e^{\frac{i\pi}n}(1-e^{\frac{2i\pi}n})\cdot e^{\frac{i\pi}n}(1-e^{\frac{4i\pi}n}) \cdots e^{\frac{i\pi}n}(1-e^{\frac{(2n-2)i\pi}n}).$$

This doesn't seem to be equal to $$\Large-\frac1ne^{\frac{i\pi}n},$$ which is what WolframAlpha gives.

share|improve this question
    
Is it $1/(z^3+1)$ or $1/(z^n+1)$? –  Julián Aguirre Nov 25 '12 at 20:21
    
@JuliánAguirre It should be $n$, sorry. I was trying do it for $n=3$ on a piece of paper and got confused. –  Bartek Nov 25 '12 at 20:22

1 Answer 1

up vote 1 down vote accepted

The easiest way is to use the following

Proposition. Let $f$ and $g$ be holomorphic functions on a neighborhood of $a\in\mathbb{C}$ such that $f(a)\ne0$ and $a$ is a simple zero of $g$. Then a is a simple pole of $f/g$ and $$ \operatorname{Res}\Bigl(\frac{f}{g},a\Bigr)=\frac{f(a)}{g'(a)}. $$ Since $e^{\tfrac{i\pi}{n}}$ is a simple zero of $1/(z^n+1)$, we get $$ \operatorname{Res}\Bigl(\frac{1}{z^n+1},e^{\tfrac{i\pi}{n}}\Bigr)=\frac{1}{n}\,e^{-\tfrac{i\pi(n-1)}{n}}. $$

share|improve this answer
    
Thank you very much! I see that this is the same as what WolframAlpha gives. But what is wrong with my solution? –  Bartek Nov 25 '12 at 20:44
    
I find your method too complicated to follow. But may be your final expression is in fact correct, and unlikely as it may seem, is equal to WolframAlpha's solution. –  Julián Aguirre Nov 25 '12 at 20:55
    
Shall I try to make my solution clearer? I would like to know if I'm doing it correctly to check if I understand the theorems. I'm very much a beginner and I constantly feel unsure that I'm doing things correctly. –  Bartek Nov 25 '12 at 21:02
    
I understand what you are doing, but I wasn't feeling like checking all the computations. I think there are some errors. Let $\omega=e^{i\pi/n}$. From $y=z-\omega$ it follows $z-e^{(2k+1)i/n}=y+\omega(1-\omega^{2k})$ and the constant term in the power series expansion of $(y+\omega(1-\omega^{2k}))^{-1}$ around $y=0$ is $\omega(1-\omega^{2k}))^{-1}$. –  Julián Aguirre Nov 26 '12 at 13:05

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.