I have a 'feeling' that the following series is equal to zero but I'm not sure how I can formally check it. The series is
$$\sum_{m \in \mathbb{Z}^2\backslash\{(0,0)\}}e^{i \theta(m)},$$
where $\theta_m$ is the argument of $m$ if m is interpreted as a complex number. For example if $m = [1, 1]$ is identified with $\hat{m} \in \mathbb{C}$ where $\hat{m} = m_1 + im_2 = 1 + i$, then $\theta(m) = \frac{\pi}{4}$.
If I think of it kind of like an integral it seems like all the oscillations should balance out and the sum should equal to zero? Or maybe this is too naive a viewpoint? How can the integral be evaluated
In this answer I will provide an arrangement such that the series $$S=\sum_{m\in\Bbb Z^2\setminus\{(0,0)\}} e^{i\theta(m)}$$converges and an arrangement such that the series diverges.
In the first case, take a rectangle $\mathcal{R}_n$ of side-length $n>1$ centered at the origin (so that $\mathcal{R}_1$ contains the points $(1,1),(1,0)(1,-1),(0,-1),(-1,-1),(-1,0)$, and $(-1,1)$). Then we have for any $n\in\Bbb N$ that $$\sum_{m\in\mathcal{R}_n}e^{i\theta(m)}=0,$$due to symmetry. Taking the limit as $n\to\infty$, we get a convergent arrangement.
For a divergent arrangement, we'll take a similar approach, but with this case, we'll take $\mathcal{R}_n$ to be a rectangle centered at the origin with side-length $n^2$; however, we'll exclude from $\mathcal{R}_n$ the points on the negative $x$-axis that are a distance of more than $n$ from the origin. This causes the points in the first, second, third, and fourth quadrants to cancel completely each time, while the positive $x$-axis contributes $n^2$ and the negative $x$-axis contributes $-n$ to the sum. Thus, we have $$\sum_{m\in\mathcal{R}_n}e^{i\theta(m)}=n^2-n.$$Clearly as $n\to\infty$, this sum tends to $\infty$.