I'm asked to evaluate the following contour integral $$\int_{\gamma} \frac{1}{(z^4+1)(z-3)}$$ where $\gamma$ is the circle of radius $2$ centered at the origin and travelled once in the counterclockwise direction.

My approach: I know how to calculate the integral but my question is, do I have to include all of the 4 singularities of $z^4+1=0$? or just one of them, say, $e^{i \pi/4}$? Or more clearly, is this integral equal to $$2 \pi i*(Res(f,3)+Res(f,e^{i \pi/4})+Res(f,e^{i3 \pi/4}) +Res(f,e^{i5 \pi/4})+Res(f,e^{i7 \pi/4}))$$ and of course, the residue at $z=3$ is zero.

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    $\begingroup$ Yes, you need to calculate residues on all singularities that contour covers. $\endgroup$ – Kaster Apr 24 '14 at 5:01
  • $\begingroup$ The residue at 3 isn't zero, it is outside the contour that interests you $\endgroup$ – vonbrand Apr 24 '14 at 6:40

Yes, you need to take into account all four singularities of $z^4 + 1$; (note also that the residues aren't equal at different poles). Perhaps it's more obvious if you write the integral as

$$\int_{\gamma} \frac{1}{(z - e^{i\pi/4})(z - e^{3i\pi/4})(z - e^{5i\pi/4})(z - e^{7i\pi/4})(z - 3)} dz $$

Also, the residue of $f$ at $z = 3$ is not zero; rather, we don't include that singularity in the sum since it doesn't lie within the contour $\gamma$.

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  • $\begingroup$ Yes pardon me. I meant to say the index is zero. $\endgroup$ – User69127 Apr 24 '14 at 5:03
  • $\begingroup$ Did u mean to put a $e^{i7 \pi/4}$ on one of your factors? $\endgroup$ – User69127 Apr 24 '14 at 5:04
  • $\begingroup$ This is such a tedious integral to compute $\endgroup$ – User69127 Apr 24 '14 at 5:12
  • $\begingroup$ Geez I think I am having difficulty computing this integral. The arithmetic is amazing. I get something very funny $\endgroup$ – User69127 Apr 24 '14 at 5:41

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