<s>[INST] Finding $\int_C \frac{8z^{11}-3z^6+1}{z^7-1}$ Find $\int_C \frac{8z^{11}-3z^6+1}{z^7-1}$ where $C$ is the positively oriented circle $|z| =4$
How should I go about finding this? We know that we have simple poles at the seventh roots of unity, i.e. $z = e^{\frac{n \pi i}{7}}$ and we also note that these poles are interior to $C: |z| =4$.
Attempt.
$$\frac{8z^{11}-3z^6+1}{z^7-1} = (8z^{11}-3z^6+1) \cdot \Big( - \frac{1}{1-z^7} \Big)$$
$$ = (8z^{11}-3z^6+1) \cdot \Big( - \displaystyle\sum_{n=0}^\infty (z^7)^n\Big) \hspace{0.5cm} |z| <1$$
$$ = (8z^{11}-3z^6+1) \cdot  \Big( -1 -z^7-z^{14} + \dots\Big) $$
But the two issues I run into here is that the geometric series is only valid for $|z|<1$ whereas we have $|z| =4$ and also I'm not seeing a $\frac{a_{-1}}{z}$ term popping out of this expansion, and I highly doubt the residue is zero. Any advice on what I should do instead?
 [/INST] It is easier to compute this integral using one resudue at infinity, due all other singularities are in interior of $C$.
$\int\limits_{|z|=4} \frac{8z^{11}-3z^6+1}{z^7-1} dz = -2 \pi i \mathop{\mathrm{Res}}_{z=\infty} \frac{8z^{11}-3z^6+1}{z^7-1} = (1)$
Using $\mathop{\mathrm{Res}}_{z=\infty} f(z) = -\mathop{\mathrm{Res}}_{z=0} \frac{1}{z^2} f\left(\frac{1}{z}\right)$ we obtain $\mathop{\mathrm{Res}}_{z=\infty} \frac{8z^{11}-3z^6+1}{z^7-1} = -\mathop{\mathrm{Res}}_{z=0}\frac{z^{11}-3 z^5+8}{z^6-z^{13}}$.
$\frac{z^{11}-3 z^5+8}{z^6-z^{13}} = z^{-6}(z^{11}-3 z^5+8)(1-z^{7})^{-1} = z^{-6}(z^{11}-3 z^5+8) \left(1 + \sum\limits_{n=1}^{\infty} z^{7 n}\right)$.
So, it is obviously that $\mathop{\mathrm{Res}}_{z=0}\frac{z^{11}-3 z^5+8}{z^6-z^{13}} = -3$. Then we have
$(1) = -2 \pi i \times (-(-3)) = -6 \pi i$.
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