Show complex summation property

[Grothard]

Homework Statement

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).

Homework Equations

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

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.

 

[mfb]

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

 

[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?

 

[mfb]

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.

 

[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.