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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?

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    $\begingroup$ Hint: The function is rather symmetric. Do you really need to take it over each pole, or can you spot a much faster pattern? $\endgroup$ Mar 28 '17 at 23:51
  • $\begingroup$ @SimplyBeautifulArt True, we can go over just three poles. $\endgroup$
    – sequence
    Mar 28 '17 at 23:53
  • $\begingroup$ You could also use the geometric series here. $\endgroup$ Mar 29 '17 at 0:00
  • $\begingroup$ I can see how to use geometric series for expansion about $0$, but what about the poles? $\endgroup$
    – sequence
    Mar 29 '17 at 0:08
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    $\begingroup$ You might want to check your factoring. Anyways, it gives you $g(z)$ for any pole $c$ in a more general form in a nice symmetric form, since $a^{n-1}b=a^{n-2}b^2=\dots$ when $a=b$, which is what you want when you asked for $g(c)$. $\endgroup$ Mar 29 '17 at 12:25
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Via this formula, the residues at each pole $z$ can be represented as $$ \operatorname{Res}(f;z) = \frac{1}{6z^5} $$

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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 $.

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