Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am in the process of computing an integral using the Cauchy residue theorem, and I am having a hard time computing the residue of a pole of high order.

Concretely, how would one compute the residue of the function $$f(z)=\frac{(z^6+1)^2}{az^6(z-a)(z-\frac{1}{a})}$$ at $z=0$?

Although it is not needed here, $a$ is a complex number with $|a|<1$.

Thanks in advance for any insight.

share|cite|improve this question
Can you compute the Taylor series at $0$? – Mariano Suárez-Alvarez Mar 19 '12 at 6:50
up vote 1 down vote accepted

You can write $$f(z) = \frac{1}{az^6} (z^{12} + 2z^6 + 1) \left(\sum_{k=0}^\infty \frac1{a^k} z^k \right) \left( \sum_{k=0}^\infty a^k z^k \right).$$ You want to extract the coefficient of $z^5$ in the product of the two series.

share|cite|improve this answer

$$g(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}}$$ we know: $$(a+b)^n =a^n+\frac{n}{1!}a^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+...+b^n$$ $$ \text{As regards }: |a|<1 $$ Taylor series of f(z) is: $$g(z)=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{1}{a-\frac{1}{a}}}{\frac{1}{a}-z}=(\frac{1}{a-\frac{1}{a}}) \left[ \frac{-\frac{1}{a}}{1-\frac{z}{a}}+\frac{a}{1-az} \right]$$ $$g(z)=(\frac{1}{a-\frac{1}{a}}) \left[ \frac{-1}{a} \sum_{n=0}^{\infty}(\frac{z}{a})^n+a \sum_{n=0}^{\infty} (az)^n \right]$$ $$f(z)=\frac{(z^6+1)^2}{az^6}g(z)=\frac{z^{12}+2z^2+1}{az^6}g(z)=\left( \frac{z^6}{a} + \frac{2}{az^4} + \frac{1}{az^6} \right)g(z)$$ $$ f(z)= \left( \frac{z^6}{a} + \frac{2}{az^4} + \frac{1}{az^6} \right) \left(\frac{1}{a-\frac{1}{a}}\right) \left[ \frac{-1}{a} \sum_{n=0}^{\infty}(\frac{z}{a})^n+a \sum_{n=0}^{\infty} (az)^n \right]$$ $$ \text{ so residue is coefficient of term }z^{-1} $$ $$ f(z)=\frac{1}{a(a-\frac{1}{a})} \left[ \frac{-1}{a}\left( \sum_{n=0}^{\infty}\frac{z^{n+6}}{a^n} +2\sum_{n=0}^{\infty}\frac{z^{n-4}}{a^n} +\sum_{n=0}^{\infty} \frac{z^{n-6}}{a^n}\right) +a \left( \sum_{n=0}^{\infty} a^nz^{n+6}+2\sum_{n=0}^{\infty} a^nz^{n-4} +\sum_{n=0}^{\infty} a^nz^{n-6} \right) \right]$$

$$ \text{residue of function at z=0 is :} $$ $$ \frac{1}{a(a-\frac{1}{a})} \left[ \frac{-1}{a}\left( 0 +2\frac{1}{a^3} +\frac{1}{a^5}\right) +a \left( 0+2a^3 +a^5 \right) \right] $$

share|cite|improve this answer

Your Answer


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.