# Residue question on contour integral

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.

• Yes, you need to calculate residues on all singularities that contour covers. – Kaster Apr 24 '14 at 5:01
• The residue at 3 isn't zero, it is outside the contour that interests you – 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 $\gamma$.
• Did u mean to put a $e^{i7 \pi/4}$ on one of your factors? – User69127 Apr 24 '14 at 5:04