Residues of $\frac{1}{z^6+1}$ without using limit

Question

One way to find residues of $f(z):=\frac{1}{z^6+1}$ is to express the function in terms of the linear multipliers of the form of $f(z)=\frac{g(z)}{z-c}$, where $c$ is a pole of $f$. For example, to find the residue at $i$ we can express $z^6+1$ in terms of its roots to obtain

$$f(z) = \frac{1/[(z+i)(z^4-z^2+1)]}{z-i}$$ and then use the formula (for a simple pole, in this case) $Res(f,i)=g(i)$, where $g(z)=1/[(z+i)(z^4-z^2+1)]$, so that the residue is $-\frac{i}{6}$.

But one would need to repeat the same process for all the poles, which is somewhat tedious. Is there a more elegant approach to finding the six residues without using the limit approach? I've thought about Laurent series expansions, but how can we find such an expansion for this function at all of the poles?

Answer 1

If $f(z) = a(z)/b(z)$ where $a(p) \ne 0$ and $b$ has a simple zero at $z=p$, the residue of $f$ there is $a(p)/b'(p)$. In this case $a(z) = 1$ and $b(z) = z^6+1$, so $b'(z) = 6 z^5$. If $\omega$ is a root of $z^6 + 1$, $\omega^5 = -1/\omega$, so the residue is $-\omega/6 $.

Answer 2

Via this formula, the residues at each pole $z$ can be represented as $$ \operatorname{Res}(f;z) = \frac{1}{6z^5} $$