I'm struggling to calculate the residue of $(z^3+1)^{-3}$ at $z=\exp(i\cdot\pi/3)$, which is the pole for the given contour. Using the formula for residue has resulted in a prohibitive double derivative (3rd order pole) and limit evaluation. I'm not sure how to produce the Laurent series for this expression either.
Any and all help is appreciated.
I don't know any particularly easy method and suspect that you just have to do the differentiation. However unless I've missed something, it's not all that bad, especially if you introduce some notation to cut down the writing, and the limit at the end only involves continuous functions so it's really just a substitution and not a limit. Let $$\alpha=e^{\pi i/3}\quad\hbox{and}\quad q(z)=z^2+\alpha z+\alpha^2\ ,$$ and note that $$\alpha^3=-1\quad\hbox{and}\quad z^3+1=(z-\alpha)q(z)\quad\hbox{and}\quad q(\alpha)=3\alpha^2\ .$$ Then the residue is $$\eqalign{ \frac{1}{2}\left.\frac{d^2}{dz^2}\frac{(z-\alpha)^3}{(z^3+1)^3}\right|_{z\to\alpha} &=\frac{1}{2}\left.\frac{d^2}{dz^2}\frac{1}{q(z)^3}\right|_{z\to\alpha}\cr &=-\frac{3}{2}\left.\frac{d}{dz}\frac{2z+\alpha}{q(z)^4}\right|_{z\to\alpha}\cr &=-\frac{3}{2}\left.\left(\frac{2}{q(z)^4}-4\frac{(2z+\alpha)^2}{q(z)^5}\right) \right|_{z\to\alpha}\cr &=-\frac{3}{(3\alpha^2)^4}+6\frac{(3\alpha)^2}{(3\alpha^2)^5}\cr &=-\frac{5\alpha}{3^3}\ .\cr}$$ Of course it's close to certain that I have made a slip somewhere here, but you get the idea ;-)
