I'd really like some help with this problem. I'm supposed to find $$ \int_{\partial B_{2}(0)} \frac{1}{(z^n-1)^2}dz,$$ where $B_2(0) = \{ z \in \mathbb{C} \; | \; |z|<2 \}$ (ie. the ball of radius 2 and centered at 0). This, of course, amounts to finding the residues. I can do it by investigating the derivative $(\text{Res}(f, z_i) = \frac{d}{dz}\left( \prod_{j \neq i} \frac{1}{(z-z_j)^2} \right)$, where I'm denoting $z_i$ the $i$-th root of unity), but it looks kind of messy attacking it straight on. Any ideas?
Also, in a later item in the same question I should be able to find a primitive for $\frac{z^4}{(1-z^3)^2}$, where $|z|<1/2$. I have absolutely no idea here (at an intermediary item, we have to find $\int_{\partial B_{1/2}(0)} \frac{z^{2n-2}}{(1-z^n)^2}$, but this is zero since there are no residues, and this calculation doesn't seem to help...). Can you give me some direction?
Thanks!
Instead of computing the sum of the residues inside your contour compute the negative of the sum of the residues outside the ball. There is only one, namely the residue at infinity, which is given by for $f(z)$ by $$ \operatorname{Res}_{z=\infty} f(z) = \operatorname{Res}_{z=0} \left(-\frac{1}{z^2} f\left(\frac{1}{z}\right)\right).$$ In the present case we have $$ -\frac{1}{z^2} f\left(\frac{1}{z}\right) = -\frac{1}{z^2} \frac{1}{((1/z)^n-1)^2} = -\frac{1}{z^2} \frac{z^{2n}}{(1-z^n)^2} = - z^{2n-2} \sum_{k=0}^\infty (k+1) z^{nk}.$$ The least power of $z$ that appears in this sum is $z^{2n-2}$ and $2n-2\ge 0$. It follows that $$\operatorname{Res}_{z=\infty} f(z) = 0$$ and the integral is zero as well.