# Homework Help: Show complex summation property

1. Mar 24, 2013

### Grothard

1. The problem statement, all variables and given/known data
Let $f(z) = \sum_{n =-\infty}^{\infty} e^{2 \pi i n z} e^{- \pi n^2}.$ Show that $f(z+i) = e^{\pi} e^{-2\pi i z}f(z).$

2. Relevant equations
Nothing specific I can think of; general complex analysis/summation techniques.

3. The attempt at a solution
$f(z+i) = \sum_{n =-\infty}^{\infty} e^{-2 \pi n z} e^{2 \pi i n z} e^{- \pi n^2};$ I can't factor the new term out of the sum because it contains an n. I feel like I might be missing some sort of summation identity that can accomplish this. I also tried completing the square, but it didn't really get me far.

2. Mar 24, 2013

### Staff: Mentor

Messing around with the exponent,
$$\exp(-2\pi n z + 2 \pi i n z - \pi n^2) = \exp(2 \pi i (n+1) z - \pi (n+1)^2 + 2\pi n + \pi - 2\pi i z - 2\pi n z) = e^\pi e^{- 2\pi i z} \exp(2 \pi i (n+1) z - \pi (n+1)^2 + 2\pi n(1-z))$$
$$e^\pi e^{- 2\pi i z} \exp(2 \pi i n z - \pi n^2 + 2\pi (n-1)(1-z))$$
which looks quite nice. But how to get rid of the last part?

3. Mar 25, 2013

### Grothard

That's really cool; it does simplify the problem a lot. But I'm having a bit of trouble getting rid of that last term. I tried shifting the index once more, but that doesn't seem to help. It can't be factored out, and it's definitely nontrivial for a given term. Could you give me some tips as to how to approach it?

4. Mar 25, 2013

### Staff: Mentor

I don't know. The approach gives the required prefactors, so it might be something into it, but I don't know how to get rid of that additional term.
It is trivial for z=1, and the function is periodic with a period of 1, therefore the equation is true for integer z.

5. Mar 25, 2013

### jashua

It seems z = 1 + ik, where k is an integer. It is also interesting to notice that f(z) is periodic with 1, as mfb has said. Then the relation between f(z) and f(z+i) must hold for z = r + ik, where both r and k are integers.

Last edited: Mar 25, 2013