Complex argument and equivalence class

We know that the complex argument of a product of two numbers is equal to the sum of their arguments. I'm aware that one can find this statement written down in many mathematics and engineering texts. However, it is a statement that assumes the argument of a complex number not to be a number but rather an equivalence class of numbers (numbers considered equivalent if they differ by a multiple of $2πi$). If I need the argument as a single number (e.g. in programming) one needs to fix a half-open interval of length $2π$ to which the argument belongs by definition. There are two popular choices: $[0,2π)$ and $(-π,π]$. For both cases it is easy to find numbers in which this statement is violated for the then uniquely defined argument.

My question is how one can avoid or resolve this problem.

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You would have to add or subtract a multiple of $2\pi$ to land in your chosen interval. –  Andrew Dec 14 '12 at 19:46

It is precisely in programming where the notion of a cyclic sum is more natural: $0xFFFF+0x0001=0x0000$.
In this case, all you need to do is define "addition modulo $2\pi$", where you subtract $2\pi$ if your sum ends up outside your desired interval.