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There are three parts of the question, the first two which I proved.

$a.$ Proving $f(z)$ is entire analytic.

$b.$ $f(z+1) = f(z)$ and $f(z+i) = e^{\pi}e^{-2\pi iz}f(z)$

$c.$ Inside the unit square in the 1st quadrant, prove that $f(z)$ has a unique $z_0$ such that $f(z_0)=0$

I need some help regarding the $c.$ part of the question. I don't understand how to apply the argument principle here (which is a hint). It is also given that $f(z)$ does not vanish on the boundary of the square.

What I could do:

$$f(0) = f(1) = 1+2\sum_{n=1}^{\infty}e^{-\pi n^2}$$

and

$$f(i) = f(1+i) = e^{\pi}\big(1+2\sum_{n=1}^{\infty}e^{-\pi n^2}\big)$$

Using the hint given below by a user: I end up with $$\dfrac{1}{2\pi i}log f(z)$$ and the integral evaluates to -1/2 and 1/2 on the two vertical sides and 0 on the horizontal lines. Resulting in the closed integral to 0. Any ideas where I might be going wrong?

Any idea how to proceed?

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    You should be able to explicitly compute the integral of $\frac{f^{\prime}}{f}$ over the boundary of the unit square using part (b). According to the argument principle, this integral is equal to the number of zeros of $f$ in the square. – carmichael561 Mar 18 '16 at 21:32
  • for $i x \in \mathbb{R}$ : $\displaystyle f(x) = e^{\textstyle-4 \pi x} \sum_n e^{-\textstyle (\sqrt{\pi} n-i \sqrt{\pi} x)^2}$ and by analytic continuation it stays true for any $x \in \mathbb{C}$ – reuns Mar 18 '16 at 21:35
  • Thanks. I spent too much time trying to use Rouche's theorem (misread the question) to figure this out. I will update once I have a solution. –  Mar 18 '16 at 21:36
  • @carmichael561 : $\frac{f'}{f}$ is a modular form plus a simple function ? – reuns Mar 18 '16 at 21:39
  • Looks like Jacobi $\vartheta$ function with $\tau = -1$. – Henricus V. Mar 18 '16 at 21:42
  • @HenryW. yes, and $\frac{f'}{f}(z+1) =\frac{f'}{f}(z) = \frac{f'}{f}(z+i)-2 i \pi$ hence $\frac{d}{dz} \frac{f'}{f}(z) = \frac{d}{dz} \frac{f'}{f}(z+1) = \frac{d}{dz} \frac{f'}{f}(z+i)$ is meromorphic and double periodic $\implies$ it is a modular form, with lattice ${1,i}$ (no smaller period since $f$ is entire and has a single zero in each period, hence $f'/f$ has single pole, as $\frac{d}{dz} \frac{f'}{f}$) – reuns Mar 19 '16 at 02:57

2 Answers2

-1

Just an idea but I'm getting zero. Perhaps you can spot something I missed.

Let $\;C_1,C_2,C_3,C_4\;$ the unit square's side begining with the one on the real axis and counterclockwise. Then, since the function is entire and non-zero on the square, we get

$$\left.\int_{C_1}\frac{f'(z)}{f(z)}dz=\log f(z)\right|_0^1=\log f(1)-\log f(0)=0$$

and the same on the other horizontal side, yet on the rightmost vertical side:

$$\int_{C_2}\frac{f'(z)}{f(z)}=\log f(1+i)-\log f(1)=\log\left(e^\pi e^{-2\pi i}f(1)\right)-\log f(1)=\pi$$

and on the other vertical side:

$$\int_{C_4}\frac{f'(z)}{f(z)}=\log f(0)-\log f(i)=\log f(0)-\log\left(e^\pi e^{0}f(0)\right)=-\pi$$

DonAntonio
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-1

$\log f(z+1) = \log f(z) \pmod {2i\pi}$ so after differentiating, $f'(z)/f(z) = f'(z+1)/f(z+1)$

$\log f(z+i) = \log (f(z)\exp(\pi-2i\pi z)) = \log (f(z)) + \pi - 2i\pi z \pmod {2i\pi}$, so after differentiating, $f'(z+i)/f(z+i) = f'(z)/f(z) - 2i\pi$.

As a result, when you integrate $f'/f dz$ along a unit square, the contributions of the vertical sides cancel each other, and the contributions of the horizontal sides give $\int_0^1 2i\pi dt = 2i\pi$.

Then, the integral of $f'/f dz$ along any unit square is $2i\pi$ (as long as $f$ doesn't vanish on the contour), and so since $f$ is entire, it must have exactly one zero inside every unit square.

Intuitively, when you look at the image of a horizontal segment of length $1$, when you move up by $i$, you multiply it by $exp(2i\pi z)$, which adds one twist around $0$ to the image of the segment by $f$.

In general, if you have two continuous functions $f,g : S^1 \to \Bbb C^*$, the winding number of $fg$ is the sum of the two windings numbers. Since you can deform both $f$ and $g$ as you want without changing the winding numbers (as long as you don't go through $0$), you can suppose that $f$ stays at $1$ on the first half of the circle, then does $n$ turns on the second half, and that $g$ stays at $1$ on the second half, then does $m$ turns on the first half. When you multiply them you obtain a function $fg$ that does $m$ turns on the first half of the circle and $n$ turns on the second half, so $n+m$ turns in total.

mercio
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