# Residue of a pole of order 6

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.

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Can you compute the Taylor series at $0$? –  Mariano Suárez-Alvarez Mar 19 '12 at 6:50

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.
$$g(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}}$$ 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: $$g(z)=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{1}{a-\frac{1}{a}}}{\frac{1}{a}-z}=(\frac{1}{a-\frac{1}{a}}) \left[ \frac{-\frac{1}{a}}{1-\frac{z}{a}}+\frac{a}{1-az} \right]$$ $$g(z)=(\frac{1}{a-\frac{1}{a}}) \left[ \frac{-1}{a} \sum_{n=0}^{\infty}(\frac{z}{a})^n+a \sum_{n=0}^{\infty} (az)^n \right]$$ $$f(z)=\frac{(z^6+1)^2}{az^6}g(z)=\frac{z^{12}+2z^2+1}{az^6}g(z)=\left( \frac{z^6}{a} + \frac{2}{az^4} + \frac{1}{az^6} \right)g(z)$$ $$f(z)= \left( \frac{z^6}{a} + \frac{2}{az^4} + \frac{1}{az^6} \right) \left(\frac{1}{a-\frac{1}{a}}\right) \left[ \frac{-1}{a} \sum_{n=0}^{\infty}(\frac{z}{a})^n+a \sum_{n=0}^{\infty} (az)^n \right]$$ $$\text{ so residue is coefficient of term }z^{-1}$$ $$f(z)=\frac{1}{a(a-\frac{1}{a})} \left[ \frac{-1}{a}\left( \sum_{n=0}^{\infty}\frac{z^{n+6}}{a^n} +2\sum_{n=0}^{\infty}\frac{z^{n-4}}{a^n} +\sum_{n=0}^{\infty} \frac{z^{n-6}}{a^n}\right) +a \left( \sum_{n=0}^{\infty} a^nz^{n+6}+2\sum_{n=0}^{\infty} a^nz^{n-4} +\sum_{n=0}^{\infty} a^nz^{n-6} \right) \right]$$
$$\text{residue of function at z=0 is :}$$ $$\frac{1}{a(a-\frac{1}{a})} \left[ \frac{-1}{a}\left( 0 +2\frac{1}{a^3} +\frac{1}{a^5}\right) +a \left( 0+2a^3 +a^5 \right) \right]$$