Show complex summation property

In summary, the function f(z) is periodic with a period of 1, and the equation between f(z) and f(z+i) holds for z = r + ik, where both r and k are integers.
  • #1
Grothard
29
0

Homework Statement


Let [itex] f(z) = \sum_{n =-\infty}^{\infty} e^{2 \pi i n z} e^{- \pi n^2}.[/itex] Show that [itex] f(z+i) = e^{\pi} e^{-2\pi i z}f(z). [/itex]


Homework Equations


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


The Attempt at a Solution


[itex] f(z+i) = \sum_{n =-\infty}^{\infty} e^{-2 \pi n z} e^{2 \pi i n z} e^{- \pi n^2};[/itex] 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.
 
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  • #2
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?
 
  • #3
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
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
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:

1. What is the complex summation property?

The complex summation property is a mathematical concept that states that the sum of two complex numbers is equal to the sum of their real parts and the sum of their imaginary parts. In other words, the real and imaginary components of complex numbers can be added separately.

2. How is the complex summation property used in science?

The complex summation property is commonly used in fields such as physics, engineering, and signal processing. It allows scientists to easily perform calculations involving complex numbers, which are often used to represent physical quantities such as electrical currents or electromagnetic waves.

3. Can the complex summation property be applied to more than two complex numbers?

Yes, the complex summation property can be applied to any number of complex numbers. The sum of multiple complex numbers is equal to the sum of their real parts and the sum of their imaginary parts.

4. Is the complex summation property commutative?

Yes, the complex summation property is commutative, meaning that the order in which the complex numbers are added does not change the result. For example, (2+3i) + (4+5i) is equal to (4+5i) + (2+3i).

5. Are there any exceptions to the complex summation property?

Yes, the complex summation property does not hold for non-commutative operations such as multiplication or division. In these cases, the real and imaginary parts must be multiplied or divided separately to get the correct result.

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