Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
Yes, sorry, I meant the boundary of the ball. Thanks for correcting my mistake, Jorge. – Peter Watts Dec 9 '12 at 5:59
Hint: Making the substitution $w=1/z$ will improve your life considerably, except possibly when $n=1$. – Micah Dec 9 '12 at 6:21
Gerry, guess I made it clear now: $B_2(0)$ is a common notation for the ball of radius 2 centered at 0. – Peter Watts Dec 9 '12 at 15:41
Micah, that really helps in connecting the integrals $\int_{\partial B_2(0)} \frac{1}{(z^n-1)^2}$ and $\int_{\partial B_{1/2}(0)}$, but I can't see why this calculation would be useful to find the value of either or to find a primitive of $\frac{z^4}{(1-z^3)^2}$... – Peter Watts Dec 9 '12 at 15:44
The integral you get after substituting ought to involve a holomorphic function which is considerably better behaved on the interior of its contour (it will be holomorphic except possibly at zero, because your original function was holomorphic on the exterior of its contour except possibly at infinity). That wasn't intended to be a hint for the second half, but if your function is poleless in $B_{1/2}(0)$ its integral is path-independent... – Micah Dec 9 '12 at 16:56

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.