Jukai

Thank you for the reply.

I will try integrating with the relation you've proposed.

As for the other method, the one I started with, the limit tends to infinity in my calculations.

If z=εe^(iθ), then dz=iεe^(iθ) and the equation you've written becomes

[tex] f(z) = \frac{1}{8} \frac{(3-4e^{2iεe^{iθ}}+e^{4iεe^{iθ}})iεe^{iθ}}{(εe^{iθ})^4} [/tex]

You can factorize your epsilon above to have ε^3 in the bottom. Now you have a form of 0/0, which is alright because you can use L'Hôpital's. After that however, we have something that tends to infinity as ε-->0.

[tex] f(z) = \frac{1}{8} \frac{(-8ie^{iθ}e^{2iεe^{iθ}}+4ie^{iθ}e^{4iεe^{iθ}})ie^{iθ}}{4ε^{3}(e^{iθ})^4} [/tex]