How to prove $f(z) = \sum_{n=-\infty}^{\infty}e^{2\pi inz}e^{-\pi n^2}$ has a unique zero inside a unit square in the 1st quadrant. 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?
 A: 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:
$$\int_{C_4}\frac{f'(z)}{f(z)}=\log f(0)-\log f(i)=\log f(0)-\log\left(e^\pi e^{0}f(0)\right)=-\pi$$
A: $\log f(z+1) = \log f(z) \pmod {2i\pi}$ 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.
