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I have to integrate $\dfrac{1}{(z-4)(z^7-1)}$ w.r.t $z$ along $C : |z|=2$.

7 poles of the function lie inside $C$ and $z=4$ lies outside. I am trying to find the answer by applying residue theorem. i.e. $2\pi i \sum (\text{residues inside }C)$, but it is becoming too lengthy to handle. Any leads?

ali mansoor
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    You can also calculate the integral by considering only those residues at points outside the contour. So you only have two summands (for $4$ and $\infty$) – leoli1 Jan 20 '21 at 14:21

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We have \begin{align} \oint_{|z|=2}f(z)\,dz&=\int_0^{2\pi}f(2e^{it})\,{2ie^{it}}\,dt =\int_0^{2\pi}f(2e^{-it})2ie^{-it}\,dt\\[0.3cm] &=\int_0^{2\pi}\frac1{\frac14\,e^{2it}}\,f\Bigg(\frac1{\frac12\,e^{it}}\Bigg)\,\frac12ie^{it}\,dt \\[0.3cm] &=\oint_{|z|=\frac12}\frac1{z^2}\,f\Big(\frac1z\Big)\,dz. \end{align} Now your new function is $$ \frac1{z^2}\,f\Big(\frac1z\Big)=\dfrac{1}{z^2\Big(\dfrac1z-4\Big)\Big(\frac1{z^7}-1\Big)} =\frac{z^6}{4\Big(z-\dfrac14\Big)(z^7-1)} $$ and you want to integrate on $|z|=\tfrac12$, so there is a single pole at $z=\tfrac14$. As the pole is simple, the integral is $$ \frac{2\pi i (1/4)^6}{4((1/4)^7-1)}=\frac{2i\pi}{1-4^7}=-\frac{2i\pi}{16383}. $$

Martin Argerami
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