<s>[INST] Help finding the residue of $1/(z^8+1)$ Help finding the residue of $1/(z^8+1)$
I'm integrating over $\{ Re^{it} | 0 \leq t \leq \pi \}$, and I found 4 simple poles at $z_0=e^{in\pi/8}$ where $n = 0,...,3$ and I'm trying to calculate $res(1/(z^8+1),z_0)$ 
calculating this: $$\lim_{z\to z_0} (z-z_0)f = \lim_{z\to z_0}\frac{z-z_0}{1+z^8},$$ now are there any algebra tricks I can do to simplify this?
 [/INST] $$z^8+1=(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})$$
$$Res_{z=\sqrt[8]i}= \lim_{x\to\sqrt[8]i}\dfrac{(z-\sqrt[8]i)}{(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})}
=\lim_{z\to\sqrt[8]i}\dfrac{(z-i^\frac{1}{8})}{(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})}
=\lim_{z\to\sqrt[8]i}\dfrac{1}{(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})}
=-\frac{i^\frac{1}{8}}{8}$$.
Now apply the same procedure to the other singularities.
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