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In my text book it says that the complex number z(not equal to 0) can be written in polar form as $z = r(\cos\theta + i \sin\theta)$, where r = mod z greater than 0 is the modulus and $\theta = \arg z$, with $-\pi \le \theta \le\pi$.

Why must the condition $-\pi < \theta \le \pi$ hold? Why not $0 < \theta \le 2 \pi$?

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    $\begingroup$ Either is acceptable: it doesn't really make any difference. It just depends on the convention adopted by the book author or exam board $\endgroup$ – David Quinn Jul 5 '15 at 9:47
  • $\begingroup$ I do not get your definition of $r$. What is $r = \mod z$? $\endgroup$ – mvw Jul 5 '15 at 10:51
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Strictly speaking, the argument of a complex number is an element of the quotient group $\mathbf R/2\pi\mathbf Z$. Usually one takes a full set of representatives of this group – mainly $(-\pi,\pi]$ and $[0,2\pi)$, i.e. a set of real numbers such that any real number is congruent, modulo $2\pi\mathbf Z$, to exactly one number in the set.

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The argument of a non-zero complex number $z$ is a multi-valued function defined by $$\arg z = \theta + 2\pi n, \, \, \, n \in \mathbb{Z}$$

where $\theta \equiv \text{Arg }z $, the principal value of the argument, which by convention is taken to be $$-\pi < \theta \leq \pi$$

But this is just by convention, you could equally well have taken the principal value to be $$0 < \theta \leq 2\pi$$ and all your work with complex numbers would be just as valid.

We now have $$\arg z = \text{Arg }z + 2\pi n, \, \, \, \, n \in \mathbb{Z}$$ we can now define $$\text{Arg }z = \arg z + 2\pi \left\lfloor\frac{1}{2} - \frac{\arg z}{2\pi}\right\rfloor $$

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