Thread: Cauchy's residue theorem - help

1. Cauchy's residue theorem - help

I'm trying to prove the reciprocity law for quadratic Gauss sums when m=1(mod 4) and need help showing something. Could anyone help? It's been a while since I done any residue calculus.

Let C be the rectangular contour with vertices, iR, -iR, 1/2m+iR and 1/2m-iR. It also has a semicircular indentation of radius epsilon (< 1) at 0.

Define

$f(z)=\frac{e^{2\pi iz^{2}/m}}{1-e^{2\pi iz}}$

Prove that

$\int_{C\:}f(z)dz=-\frac{1}{2}\overset{m-1}{\underset{r=1}{\Sigma}}e^{2\pi ir^{2}/m}$

Clearly f is analytic everywhere, except for a first order pole at each integer, but I am stumped at how to get the required result.

2. It looks like a relatively straight-forward application of the residue theorem. You need to find all the poles inside your rectangle, and then evaluate the residues of f(z) at each of those poles. Can you go from here?

3. Okay.

I think its the fact that m=1(mod 4) that is throwing me.

When m=1, there are no poles in the rectangle and so the integral is zero. For m=5, there is a pole at 1 and 2. At m=9 there are poles at 1, 2, 3 and 4 and so on.

Do I need to continue in this way for infinitely many poles, or have I got the wrong idea here?

4. Think I've just sent myself on a wild goose chase with that idea.

Where are the poles inside this rectangle, since they vary with different m, correct?

5. The poles of $f$ occur precisely when $1=e^{2\pi i z}.$ The numerator is never zero, so we don't have any zero-over-zero limits to worry about. Using Euler's formula, we have

$1=\cos(2\pi z)+i\sin(2\pi z).$

For this equation to hold, $\sin(2\pi z)=0,$ and $\cos(2\pi z)=1.$ This occurs, as you have said, precisely when $z\in\mathbb{Z}.$

Now, if I examine the contour you've chosen, there is one question I would ask: the indent at the origin: is that to the right or to the left? If it's to the right, then I would argue that the integral $\oint_{C}f(z)\,dz=0,$ contrary to the hypothesis. All the poles are integers. If I look at the real axis and scan the interval $[0+\varepsilon,1/2m]$ for integers $m$, I don't think there are any in that region, and therefore no poles in that region. Therefore, the contour integral of the analytic function around a simple closed contour is zero, by the Cauchy-Goursat theorem.

Are you sure you've stated everything in the problem accurately?

6. Everything you have mentioned is correct. The indent is to the right and so excludes 0.

I'm working from Apostols "Analytic Number Theory", where he discusses the transformation f(z+1). If I use this, then the zero pole would be inside the contour, but would still not give the result.

7. Can you give me an exact phrase near where you're working in Apostol? That way, I can find where you are on Amazon's previewer.

8. Page 196 - just under Thm 9.16

9. Hmm. I can't view that page at Amazon. I think someone else is going to have to take over for me here. Maybe Plato or Opalg can help you. Send a PM to them and see if they'll help out.

10. Okay.

11. You're welcome. Hope you figure it out.

12. Originally Posted by Cairo
I'm trying to prove the reciprocity law for quadratic Gauss sums when m=1(mod 4) and need help showing something. Could anyone help? It's been a while since I done any residue calculus.

Let C be the rectangular contour with vertices, iR, -iR, 1/2m+iR and 1/2m-iR. It also has a semicircular indentation of radius epsilon (< 1) at 0.

Define

$f(z)=\frac{e^{2\pi iz^{2}/m}}{1-e^{2\pi iz}}$

Prove that

$\int_{C\:}f(z)dz=-\frac{1}{2}\overset{m-1}{\underset{r=1}{\Sigma}}e^{2\pi ir^{2}/m}$

Clearly f is analytic everywhere, except for a first order pole at each integer, but I am stumped at how to get the required result.
The poles inside the contour are at $z=k$, for $1\leqslant k\leqslant (m-1)/2$. To find the residue at z=k, use the formula $\displaystyle\text{Res}(f,k) = \lim_{z\to k}(z-k)f(z) = \lim_{z\to k}\frac{z-k}{1-e^{2\pi iz}}e^{2\pi iz^{2}/m}$. Check using l'Hôpital's rule that this limit is $ie^{2\pi ik^2/m}/(2\pi)$.

Therefore by Cauchy's theorem the integral round the contour is equal to $\displaystyle\sum_{k=1}^{\frac12(m-1)}-e^{2\pi ik^2/m}$.

To get that result into the form required, notice that $e^{2\pi i(m-k)^2/m} = e^{2\pi ik^2/m}$. (The reason for that is that $(m-k)^2/m = m-2k+k^2/m$, and e to the power $2\pi i$ times an integer is equal to 1.) Thus the term $e^{2\pi ik^2/m}$ in the above sum can be split into two, as $\frac12\bigl(e^{2\pi ik^2/m} + e^{2\pi i(m-k)^2/m}\bigr)$, and then you can rearrange the sum so that it goes from 1 to m–1 instead of from 1 to (1/2)(m–1).