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?
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:

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

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