<s>[INST] Tricky integral in a complex plane How to integrate using residues:
$$\oint\limits_{|z|=2}\frac{1}{(z^6-1)(z-3)} dz $$
if the idea probably requires to change the sum of 6 residues to aditive inverse of residue at infinity.
I believe that the sum of 6 residues is:
$$2 \pi i\sum_{i=1}^6 Res_i \frac{1}{(z^6-1)(z-3)}=-2\pi i(Res_\inf \frac {1}{(z^6-1)(z-3)}+Res_3 \frac {1}{(z^6-1)(z-3)})$$
because $${3>|z|=2, \infty>|z|=2.}$$
Since $$\frac {1}{(z^6-1)(z-3)} = \frac {1}{(z^7)(1-\frac{1}{z^6})(1-\frac {3}{z})} = \frac {1}{z^7}+ \frac {3}{z^8}+ ...$$
$$=> -2\pi iRes_\inf \frac {1}{(z^6-1)(z-3)}=0$$
 [/INST] $$Res_{\infty}f(z)=Res_{z=0}-\frac{1}{z^2}f(\frac1z)=Res_{z=0}\frac{z^5}{(z^6-1)(1-3z)}=0$$
$$Res_{z=3}f(z)=Res_{z=3}\frac{1}{(z^6-1)(z-3)}=\left.\frac{1}{z^6-1}\right\rvert_{z=3}=\frac{1}{728}.$$
$$I=-2\pi i(Res_{\infty}f(z)+Res_{z=3}f(z))=-\frac{\pi i}{364}.$$
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