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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

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    How do you do the summation? The series is not absolutely convergent, so you'll need to choose an order to sum in. – 2017-02-27
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    $\left|e^{i\theta(m)}\right|=1$ doesn't converge to $0$, so your series is divergent. – 2017-02-27
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    @AdamHughes I don't know how to order the summation. Is there a natural ordering that can be taken? This series arose while I was working with a periodic lattice in $\mathbb{R}^2$. $m \in \mathbb{R}^2$ essentially represents the location of an 'object'..and there is an infinite grid of objects, hence the summation. I really don't know how you would go about choosing an ordering though. But there are a lot of periodic problems in math and physics so it seems like there should be some natural ordering for this type of problem? – 2017-02-27
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    @csss this series (on its face) diverges because the function isn't absolutely summable over the set you want. If you demand that this be a limiting process, over some kind of symmetric, bounded set like a rectangle, then each of the partial sums can be organized to be $0$ as needed, but that sum does not converge as written. – 2017-02-27
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    @AdamHughes Is this like what user420255 is suggesting..consider a finite domain where the terms cancel each other out, and then take the limit as the domain becomes infinitely large? How rigorous is this? ..by demanding that this be a limiting process are there any side effects? It seems 'too good to be true' that a series that diverges can be made to converge! – 2017-02-27
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    It's the the same as what you write @csss limits vs absolutely summable series have different interpretations. If your definition is as a limit, you're fine, but if not you are very not fine. If this was a problem from a professor, I would check with what they mean and indicate the issues we brought up here and see if he/she can clarify. – 2017-02-27
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    @csss And your intuition is wrong on convergent vs divergent: Take the alternating harmonic series, there are ways of rearranging it so that you can get any number you want as the sum! That's why absolute convergence is so important, or having such things defined by a specific limiting process, otherwise the "values" you assign the sums are meaningless. – 2017-02-27
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    @AdamHughes This wasn't from a professor, it arose while I experimenting with an expression involving a periodic lattice to see what happens under certain conditions. – 2017-02-27
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    @user420255: You have to be careful: what are the partial sums here? In a sense, a sequence has to be defined first before you consider partial sums unless a series is absolutely convergent. In your first comment, you take the partial sums to be those terms contained with a given radius, but this is NOT the only possible interpretation. – 2017-02-27
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    @csss then there is no a priori authoritative answer here. That series diverges by the usual generalized definition of an infinite sum, unless you know that there is a good reason to pretend it is a sum over balls (some contexts make sense, but yours doesn't sound like it since it's supposed to be a lattice invariant of sorts) – 2017-02-27
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    @Clayton: Thank you. I understand my mistake. I'll delete all my comments. – 2017-02-28

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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 $\infty$.

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    Thanks for demonstrating the issue with the summation. Is there a 'natural' order that can be taken? If a series like this arises in the course of some professional mathematicians work how would he/she deal with it? It seems like you can just take the ordering that makes it go to zero, or infinity, depending on which you prefer..and make things work out..which seems unrigorous! – 2017-02-27
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    @csss: Supposing a professional mathematician came across such a sum, the most he/she could do is take it as a finite sum in some way and explain precisely how you're taking the sum (e.g., when you want to consider how divergent sums behave asymptotically, you look at a finite version and estimate it). What comes to mind with the comment in parentheses is estimating $\sum\frac1n$. This is well-known to be divergent, but if you look at the finite sums with terms up to $x$, you can show $\sum_{n\leq x}\frac1n\sim\ln x$. The important thing is that how the finite terms are taken is understood. – 2017-02-28